Similar Triangles and Area Ratios

Competitive Mathematics · SSC CGL

5 Part46 worked examples125 practice questions

When you enlarge or shrink a photo the shape stays the same; every length is simply multiplied by one number. In geometry this is called similarity. Two triangles are similar when their angles are equal and the ratio of every pair of matching sides is the same. Almost every exam question about ratios of lengths, heights, shadows or areas runs on this one idea behind the scenes.

First we learn to spot similarity, then questions with parallel lines, the altitude on the hypotenuse, the trapezium, ratio chains and bow-ties, and finally the square rule for areas. The area ratios that do not need similarity (same height, midpoints, parallelograms) are in the next chapter, “Triangle Area Ratios”.

What you see	Which method
two equal angles / “show they are similar”	Same shape, matching sides
$D E ∥ BC$ or three parallel lines	Thales’ theorem
altitude on the hypotenuse of a right triangle	Altitude on the hypotenuse
a parallel line in a trapezium	The trapezium
several lines cutting each other, a ratio asked	Drawing a parallel
two poles / ladders crossing, a bow-tie	Bow-tie and crossed ladders
shadow, pole, lamp, mirror	Shadows and poles
a $tan$ or $sin$ value is given	Trigonometric similarity
a ratio of areas and similar triangles	The square rule

Start hereFirst job in every question: draw the figure and mark equal angles (matching arcs). Second job: redraw the two similar triangles separately, in the same orientation. Most mistakes come from matching the wrong sides.

Every method follows one order: the idea, the formula, how to recognise it, steps, solved examples, then questions to do yourself (open a solution only after trying). The figures play by themselves: lines draw in and angles appear one after another, in the order of the argument.

The basics of similarity

AA · matching the sides

What similar triangles are, which quantities scale together and how to match the corresponding sides.

Same shape, one multiplier

The AA test · what scales

Two equal angles make the triangles similar; then every length changes by the same multiplier.

The idea

Two triangles are similar ( $\sim$ ) if all three angles are equal. They then have the same shape, one just bigger. The convenience: you need not check all three angles. If two angles are equal the third is automatically equal (subtract from $18 0^{\circ}$ ). This is the AA test (angle-angle).

◭ Diagram

△ A BC \sim △ PQR

: angles

α, β, γ

equal, every side

\times 1.5

(blue/green = the two triangles; one arc colour = one equal angle)

Sides opposite equal angles are corresponding. Here $BC$ is opposite $∠ A$ and $QR$ is opposite $∠ P$ , so they correspond. The multiplier (ratio) is the same for every pair:

∑ Formula

\frac{PQ}{A B} = \frac{QR}{BC} = \frac{RP}{C A} = k

Similar triangles: ratio of corresponding sides

$k$: multiplier (scale factor); here $k = 1.5$
$A B, BC, C A$: sides of the first triangle
$PQ, QR, RP$: the matching sides of the second triangle

Many other things change by the same $k$ ; the exam loves this:

◭ Diagram

The altitude (green), median (blue) and angle bisector (orange) of the bigger triangle are also

\times 1.5

Quantity	Ratio	Remember
sides, perimeter	$k$	sums scale too
median, altitude, bisector	$k$	pair the matching ones
inradius, circumradius	$k$	any length
angles	1	never change
area	$k^{2}$	the square rule (later in this chapter)

Here: $k$ = the multiplier; “length” means any straight measure (a side, a line, a radius).

Watch outSimilarity means equal ratios of sides and equal angles. Differences or sums of sides are never kept equal. When writing ratios keep the same triangle on top every time: always (second triangle)/(first triangle), or always the reverse. Switching halfway ruins everything.

Steps

Find equal angles (parallel lines, right angles, a shared angle, vertically opposite angles) and mark them with matching arcs.
With two equal angle pairs the triangles are similar. Write the similarity statement in the order of the equal angles: $△ A BC \sim △ PQR$ .
Write the chain of ratios of corresponding sides: $\frac{A B}{PQ} = \frac{BC}{QR} = \frac{C A}{RP}$ .
Pick only the two ratios that hold the unknown and the given numbers, and cross-multiply.

Example 1 · Basic

△ A BC \sim △ PQR

with

A B = 14

cm,

BC = 21

cm,

C A = 28

cm and

PQ = 10

cm. Find

QR

and

RP

Solution, step by step

1The letter order gives $A \leftrightarrow P, B \leftrightarrow Q, C \leftrightarrow R$ . So $PQ$ matches $A B$ , $QR$ matches $BC$ and $RP$ matches $C A$ .
2To get the multiplier take the pair whose two numbers are both given: $k = \frac{PQ}{A B} = \frac{10}{14} = \frac{5}{7}$ .
3 $QR = k \times BC = \frac{5}{7} \times 21 = 15$ cm.
4 $RP = k \times C A = \frac{5}{7} \times 28 = 20$ cm.

Answer $QR = 15$ cm, $RP = 20$ cm

Check:

10 : 15 : 20 = 2 : 3 : 4

and

14 : 21 : 28 = 2 : 3 : 4

; same ratio.

Example 2 · Standard

△ A BC \sim △ D EF

. The perimeter of

△ A BC

is 36 cm and of

△ D EF

is 54 cm. If the median

A M

△ A BC

is 8 cm, what is the corresponding median

D N

△ D EF

Solution, step by step

1The perimeter is a sum of lengths, so the ratio of perimeters is the multiplier: $k = \frac{54}{36} = \frac{3}{2}$ .
2A median is also a length, and $A \leftrightarrow D$ gives $A M \leftrightarrow D N$ . So $D N = k \times A M$ .
3 $D N = \frac{3}{2} \times 8 = 12$ cm.

Answer 12 cm

No sides were given, yet the perimeters gave the multiplier. That is the power of similarity.

Example 3 · Standard

The corresponding sides of two similar triangles are in the ratio

3 : 5

. The difference of their perimeters is 18 cm. Find both perimeters and, if the larger triangle’s altitude is 15 cm, the smaller one’s altitude.

Solution, step by step

1The perimeters are also in the ratio $3 : 5$ . Let the smaller be $3 u$ and the larger $5 u$ .
2The difference is $5 u - 3 u = 2 u = 18$ , so $u = 9$ .
3Perimeters: $3 \times 9 = 27$ cm and $5 \times 9 = 45$ cm.
4Altitudes are also in the ratio $3 : 5$ : the smaller one is $15 \times \frac{3}{5} = 9$ cm.

Answer Perimeters 27 cm and 45 cm; the smaller altitude is 9 cm

In ratio-with-sum/difference questions take a “unit”

u

Example 4 · Advanced

△ A BC

∠ A = 5 0^{\circ}

∠ B = 6 0^{\circ}

BC = 8

cm,

A B = 10

cm,

A C = 7

cm. In

△ PQR

∠ Q = 6 0^{\circ}

∠ R = 7 0^{\circ}

and

QR = 12

cm. Find

PQ

and

PR

Solution, step by step

1Find the third angles: $∠ C = 180 - 50 - 60 = 7 0^{\circ}$ and $∠ P = 180 - 60 - 70 = 5 0^{\circ}$ .
2Match the angles: $A = 5 0^{\circ} = P$ , $B = 6 0^{\circ} = Q$ , $C = 7 0^{\circ} = R$ . So $△ A BC \sim △ PQR$ (AA).
3The corresponding side is the one opposite the equal angle: $BC$ is opposite $∠ A$ and $QR$ is opposite $∠ P$ . So $k = \frac{QR}{BC} = \frac{12}{8} = \frac{3}{2}$ .
4 $A B$ and $PQ$ are opposite $∠ C = ∠ R = 7 0^{\circ}$ : $PQ = \frac{3}{2} \times 10 = 15$ cm.
5 $A C$ and $PR$ are opposite $∠ B = ∠ Q$ : $PR = \frac{3}{2} \times 7 = 10.5$ cm.

Answer $PQ = 15$ cm, $PR = 10.5$ cm

When only angles are given, match by the opposite side; do not trust the letter order.

Try it yourself

◭ Diagram

BasicTry it yourself first

△ A BC \sim △ D EF

with

A B = 12

cm,

D E = 18

cm and

BC = 20

cm. Find

EF

◭ Diagram

BasicTry it yourself first

△ PQR \sim △ X Y Z

with

PQ = 15

cm,

X Y = 9

cm and

QR = 25

cm. Find

Y Z

BasicTry it yourself first

Two similar triangles have corresponding sides 6 cm and 9 cm. The first has perimeter 36 cm. Find the perimeter of the second.

BasicTry it yourself first

Two similar triangles have corresponding sides 8 cm and 20 cm. A median of the first triangle is 6 cm. Find the corresponding median of the second.

StandardTry it yourself first

Two similar triangles have corresponding sides 9 cm and 15 cm. An altitude of the first triangle is 12 cm. Find the corresponding altitude of the second.

StandardTry it yourself first

The corresponding altitudes of two similar triangles are in the ratio

4 : 7

and their sum is 33 cm. Find the larger altitude.

StandardTry it yourself first

△ A BC

∠ A = 4 0^{\circ}

∠ B = 7 5^{\circ}

BC = 9

cm and

C A = 12

cm. In

△ X Y Z

∠ Y = 4 0^{\circ}

∠ Z = 6 5^{\circ}

and

ZX = 15

cm. Find

Y Z

StandardTry it yourself first

Two similar triangles have perimeters 40 cm and 64 cm. The longest side of the smaller is 15 cm. Find the longest side of the larger.

AdvancedTry it yourself first

A triangle has sides 9, 12 and 15 cm and inradius (radius of the inner circle) 3 cm. A similar triangle has inradius 2 cm. Find its perimeter.

AdvancedTry it yourself first

The triangle with sides 6, 8, 10 cm has circumradius (radius of the outer circle) 5 cm, because it is right-angled and the radius is half the hypotenuse. A similar triangle has longest side 25 cm. Find its circumradius.

Matching the corresponding sides

Three forms · redraw apart

Recognising similarity by AA, SAS, SSS and the equal-angle (antiparallel) form, and matching the sides correctly.

The idea

In a crowded figure the two similar triangles sit inside each other, so the matching goes wrong. The cure is one habit: redraw the two triangles apart, in the same orientation. Then equal angles sit in the same place and the corresponding sides pair up by themselves.

There are three tests for similarity (any one is enough):

Test	What to check	When it helps
AA	two angles equal	parallel lines, right angles, a shared angle
SAS	one angle equal and the two sides around it in the same ratio	a shared angle but no parallel line
SSS	all three side ratios equal	only lengths are given

Here: AA = angle-angle; SAS = side-angle-side (the angle lies between the two sides); SSS = side-side-side (all three sides).

A very useful form: the equal-angle (antiparallel) line

In $△ A BC$ let $D$ lie on $A B$ and $E$ on $A C$ with $∠ A D E = ∠ A CB$ (not parallel, but a crossed equal angle). Then $△ A D E \sim △ A CB$ , because angle $A$ is shared and one more pair of angles is equal. Notice: $D$ matches $C$ , and $E$ matches $B$ .

◭ Diagram

Left:

∠ A D E = ∠ A CB

(orange arcs). Right: the two triangles pulled apart:

A \leftrightarrow A, D \leftrightarrow C, E \leftrightarrow B

∑ Formula

\frac{A D}{A C} = \frac{A E}{A B} = \frac{D E}{CB} ⟹ A D \times A B = A E \times A C

The equal-angle form (

∠ A D E = ∠ A CB

)

$A D, A E$: sides of the small triangle $A D E$
$A C, A B$: the matching sides of the big triangle
$D E, CB$: the third pair

Product shortcut: $A D \times A B = A E \times A C$ : the two lengths on the same line ( $A D$ and $A B$ both on line $A B$ ) multiply to the product of the two lengths on the other line.

Another version of the same idea: $∠ A C D = ∠ A BC$ ( $D$ on $A B$ ). Then $△ A C D \sim △ A BC$ (angle $A$ shared) and $A C^{2} = A D \times A B$ .

Steps

Mark the angles said to be equal, and the shared angle, with arcs.
Redraw the two triangles apart with the shared-angle vertex in the same place.
Write the matching from the order of angles ( $A \to A, D \to C, E \to B$ ) and the chain of ratios.
Pick two ratios and cross-multiply.

Example 5 · Standard

△ A BC

D

lies on

A B

and

E

A C

with

∠ A D E = ∠ A CB

. If

A D = 8

cm,

A B = 18

cm and

A E = 9

cm, find

A C

and

EC

Solution, step by step

1Angle $A$ is shared by $A D E$ and $A CB$ and $∠ A D E = ∠ A CB$ . By AA, $△ A D E \sim △ A CB$ ( $A \to A, D \to C, E \to B$ ).
2Corresponding sides: $A D \leftrightarrow A C$ and $A E \leftrightarrow A B$ . So $\frac{A D}{A C} = \frac{A E}{A B}$ .
3Substitute: $\frac{8}{A C} = \frac{9}{18} = \frac{1}{2}$ , so $A C = 16$ cm.
4 $EC = A C - A E = 16 - 9 = 7$ cm.

Answer $A C = 16$ cm, $EC = 7$ cm

Shortcut:

A D \times A B = A E \times A C

⟹

8 \times 18 = 9 \times A C

⟹

A C = 16

Example 6 · Standard

△ A BC

D

is on

A B

∠ A C D = ∠ A BC

A D = 4

cm and

D B = 5

cm. Find

A C

Solution, step by step

1Triangles $A C D$ and $A BC$ share angle $A$ and $∠ A C D = ∠ A BC$ . So $△ A C D \sim △ A BC$ ( $A \to A, C \to B, D \to C$ ).
2Corresponding sides: $A C \leftrightarrow A B$ and $A D \leftrightarrow A C$ . So $\frac{A C}{A B} = \frac{A D}{A C}$ .
3Cross-multiplying: $A C^{2} = A D \times A B = 4 \times 9 = 36$ ( $A B = A D + D B = 9$ ).
4 $A C = 6$ cm.

Answer 6 cm

A C

appears in both triangles, which is why a square came:

A C

is the “middle term”.

Example 7 · Standard

△ A BC

and

△ PQR

∠ A = ∠ P

A B = 6

cm,

A C = 9

cm,

PQ = 8

cm and

PR = 12

cm. If

BC = 12

cm find

QR

Solution, step by step

1The sides around $A$ are $A B, A C$ and around $P$ are $PQ, PR$ . Check the ratios: $\frac{PQ}{A B} = \frac{8}{6} = \frac{4}{3}$ and $\frac{PR}{A C} = \frac{12}{9} = \frac{4}{3}$ .
2Both are equal and the included angle is equal, so by SAS $△ A BC \sim △ PQR$ with $k = \frac{4}{3}$ .
3The third pair: $QR = \frac{4}{3} \times BC = \frac{4}{3} \times 12 = 16$ cm.

Answer 16 cm

In SAS the angle must be between the two sides, or the test does not apply.

Example 8 · Basic

Three triangles have sides

X

: 4, 6, 8;

Y

: 6, 9, 12;

Z

: 5, 7, 9. Which are similar?

Solution, step by step

1To test SSS put the sides of each triangle in order and divide by the smallest.
2Write each triple as a simplest ratio: $X$ : $4 : 6 : 8 = 2 : 3 : 4$ ; $Y$ : $6 : 9 : 12 = 2 : 3 : 4$ ; $Z$ : $5 : 7 : 9$ , which is not $2 : 3 : 4$ .
3 $X$ and $Y$ have equal ratios ( $k = \frac{3}{2}$ ), so $X \sim Y$ ; $Z$ is similar to neither.

Answer Only $X \sim Y$

When testing ratios compare shortest↔shortest, longest↔longest.

Try it yourself

◭ Diagram

BasicTry it yourself first

△ A BC

D

is on

A B

and

E

A C

with

∠ A D E = ∠ A CB

. If

A D = 6

D B = 8

and

A E = 7

(in cm), find

EC

◭ Diagram

StandardTry it yourself first

△ A BC

D

is on

A B

and

E

A C

with

∠ A D E = ∠ A CB

. If

A D = 9

D B = 11

and

A E = 12

(in cm), find

EC

◭ Diagram

BasicTry it yourself first

△ A BC

D

is on

A B

and

∠ A C D = ∠ A BC

. If

A D = 4

cm and

D B = 5

cm, find

A C

◭ Diagram

StandardTry it yourself first

△ A BC

D

is on

A B

and

∠ A C D = ∠ A BC

. If

A D = 9

cm and

D B = 16

cm, find

A C

StandardTry it yourself first

△ A BC

and

△ D EF

∠ B = ∠ E

A B = 10

BC = 15

D E = 14

EF = 21

(cm). If

A C = 12

, find

D F

StandardTry it yourself first

About triangles with sides

(3, 5, 7)

and

(6, 10, 15)

, which statement is correct?

AdvancedTry it yourself first

△ A BC

D

is on

A B

and

E

A C

with

∠ A E D = ∠ A BC

. If

A D = 6

cm,

A C = 15

cm and

BC = 20

cm, find

D E

Parallel lines and ratios

Thales · trapezium · chains · bow-tie

When one line is parallel to another everything divides in proportion; this solves trapezium, ratio-chain and bow-tie questions.

Thales’ theorem (basic proportionality)

A parallel line divides the sides in proportion

A line parallel to a side divides the other two sides in the same ratio; the converse and three parallel lines.

The idea

Draw a line $D E$ parallel to $BC$ in $△ A BC$ ( $D$ on $A B$ , $E$ on $A C$ ). Parallel lines give equal corresponding angles: $∠ A D E = ∠ A BC$ and $∠ A E D = ∠ A CB$ . Angle $A$ is shared, so $△ A D E \sim △ A BC$ (AA). Similarity then gives $\frac{A D}{A B} = \frac{A E}{A C} = \frac{D E}{BC}$ . This is Thales’ theorem, and its most useful form is:

◭ Diagram

D E ∥ BC

: upper parts (blue) 2 : 2 and lower parts (green) 3 : 3 — the same ratio on both sides

∑ Formula

\frac{A D}{D B} = \frac{A E}{EC} and \frac{A D}{A B} = \frac{A E}{A C} = \frac{D E}{BC}

Thales’ theorem (

D E ∥ BC

)

$A D, D B$: the two pieces of side $A B$ (split at $D$ )
$A E, EC$: the two pieces of side $A C$ (split at $E$ )
$A B, A C$: the whole sides
$D E, BC$: the parallel segment and the base it matches

Watch outTwo kinds of ratio, both correct: (1) part : part,

\frac{A D}{D B} = \frac{A E}{EC}

; (2) part : whole,

\frac{A D}{A B} = \frac{A E}{A C} = \frac{D E}{BC}

. The classic trap: setting

\frac{D E}{BC}

equal to

\frac{A D}{D B}

D E : BC

always goes with the part : whole ratio.

The converse: is the line parallel?

The reverse is true too: if a line divides two sides in the same ratio ( $\frac{A D}{D B} = \frac{A E}{EC}$ ) it is parallel to the third side. This solves “show that $D E ∥ BC$ ” questions: compute both ratios and compare.

◭ Diagram

Left: 2 : 3 and 2 : 3 are equal, so the line is parallel (green, ✓). Right: 2 : 3 and 5 : 3 are not equal, so the line is tilted (pink, ✗)

Three parallel lines

Three (or more) parallel lines cut any two transversals in the same ratio. This is Thales in a bigger setting.

◭ Diagram

l_{1} ∥ l_{2} ∥ l_{3}

: blue piece : green piece is the same on both transversals

∑ Formula

\frac{A B}{BC} = \frac{D E}{EF}

Three parallel lines

$A B, BC$: consecutive pieces of the first transversal
$D E, EF$: the matching pieces of the second transversal

Steps

Find which line is parallel to which side (from the question or the figure).
Name the given and asked pieces on both sides: $A D, D B$ on one side; $A E, EC$ on the other.
If two parts are given use part : part; if a whole side or $BC$ is given use part : whole.
Cross-multiply to get the unknown.

Example 9 · Basic

△ A BC

D E ∥ BC

with

D

A B

and

E

A C

A D = 6

cm,

D B = 9

cm,

A E = 8

cm. Find

EC

Solution, step by step

1 $D E ∥ BC$ , so Thales applies: on both sides upper part : lower part is equal.
2 $\frac{A D}{D B} = \frac{A E}{EC}$ , that is $\frac{6}{9} = \frac{8}{EC}$ .
3Cross-multiplying: $6 \times EC = 9 \times 8 = 72$ , so $EC = 12$ cm.

Answer $EC = 12$ cm

Check:

\frac{6}{9} = \frac{2}{3}

and

\frac{8}{12} = \frac{2}{3}

. ✓

Example 10 · Standard

△ A BC

D E ∥ BC

with

A D = (x + 1)

cm,

D B = (2 x - 1)

cm,

A E = (x - 1)

cm and

EC = (3 x - 9)

cm. Find

x

Solution, step by step

1Thales: $\frac{A D}{D B} = \frac{A E}{EC}$ ⟹ $\frac{x + 1}{2 x - 1} = \frac{x - 1}{3 x - 9}$ .
2Cross-multiply: $(x + 1) (3 x - 9) = (2 x - 1) (x - 1)$ , i.e. $3 x^{2} - 6 x - 9 = 2 x^{2} - 3 x + 1$ .
3Bring everything to one side: $x^{2} - 3 x - 10 = 0$ , i.e. $(x - 5) (x + 2) = 0$ , so $x = 5$ or $x = - 2$ .
4 $x = - 2$ would make $A E = x - 1 = - 3$ , which cannot be a length. So $x = 5$ .
5Check: $A D = 6, D B = 9, A E = 4, EC = 6$ and $\frac{6}{9} = \frac{4}{6}$ . ✓

Answer $x = 5$

When the quadratic gives two roots keep the one that makes every length positive.

Example 11 · Standard

Three parallel lines

l_{1}, l_{2}, l_{3}

cut two transversals at

A, B, C

and

D, E, F

. If

A B = 6

BC = 9

and

D F = 35

cm, find

D E

and

EF

Solution, step by step

1Three parallel lines cut both transversals in the same ratio: $D E : EF = A B : BC = 6 : 9 = 2 : 3$ .
2The whole $D F = D E + EF = 35$ cm is $2 + 3 = 5$ parts, so 1 part $= \frac{35}{5} = 7$ cm.
3 $D E = 2 \times 7 = 14$ cm and $EF = 3 \times 7 = 21$ cm.

Answer $D E = 14$ cm, $EF = 21$ cm

When the whole length is given turn the ratio into “parts” and share.

Example 12 · Advanced

△ A BC

D

is on

A B

E

A C

A D = 4

cm,

D B = 6

cm,

A E = 6

cm,

EC = 9

cm and

BC = 15

cm. Is

D E ∥ BC

? If so, find

D E

Solution, step by step

1Test the converse: $\frac{A D}{D B} = \frac{4}{6} = \frac{2}{3}$ and $\frac{A E}{EC} = \frac{6}{9} = \frac{2}{3}$ . They are equal, so $D E ∥ BC$ .
2For $D E$ we need part : whole: $A B = A D + D B = 10$ , so $\frac{D E}{BC} = \frac{A D}{A B} = \frac{4}{10} = \frac{2}{5}$ .
3 $D E = \frac{2}{5} \times 15 = 6$ cm.
4The wrong way: $\frac{D E}{15} = \frac{4}{6}$ would give 10, which is wrong.

Answer Yes, $D E ∥ BC$ and $D E = 6$ cm

D E : BC

always goes with the whole-side ratio (

A B

A C

Try it yourself

◭ Diagram

BasicTry it yourself first

△ A BC

D

is on

A B

and

E

A C

with

D E ∥ BC

. If

A D = 4

cm,

D B = 6

cm and

A E = 10

cm, find

EC

◭ Diagram

BasicTry it yourself first

△ A BC

D

is on

A B

and

E

A C

with

D E ∥ BC

. If

A D = 9

cm,

D B = 15

cm and

A E = 12

cm, find

EC

◭ Diagram

BasicTry it yourself first

△ A BC

D E ∥ BC

(

D

A B

E

A C

). If

A B = 15

cm,

A D = 6

cm and

A C = 20

cm, find

A E

◭ Diagram

StandardTry it yourself first

△ A BC

D E ∥ BC

and

A D : D B = 3 : 2

. If

BC = 25

cm, find

D E

StandardTry it yourself first

△ A BC

D E ∥ BC

with

A D = (x + 2)

cm,

D B = (x + 4)

cm,

A E = (x - 2)

cm and

EC = (x - 1)

cm. Find

x

StandardTry it yourself first

Three parallel lines cut a transversal at

A, B, C

with

A B = 8

cm and

BC = 12

cm. The same lines cut a second transversal so that

D F = 50

cm in all. How long is

EF

StandardTry it yourself first

Points

D

(on

A B

) and

E

(on

A C

) are chosen in

△ A BC

. In which case is

D E ∥ BC

proved?

AdvancedTry it yourself first

△ A BC

D

is the midpoint of

A B

and

D E ∥ BC

(

E

A C

). The perimeter of

△ A BC

is 54 cm. Find the perimeter of

△ A D E

A parallel line in a trapezium

Any line · mid-segment · diagonals

The length of a line drawn between the parallel sides: a general line, the mid-segment, the diagonal midpoints and the line through the crossing point.

The idea

A trapezium has one pair of parallel sides (the bases), here $A B = a$ at the bottom and $D C = b$ at the top. Draw any line $EF$ parallel to the bases between the non-parallel sides ( $A D$ and $BC$ ); its length lies between $a$ and $b$ . One trick finds it: draw the diagonal $D B$ . Two triangles appear and Thales works in both.

◭ Diagram

EF ∥ A B

and the diagonal

D B

meets it at

M

: blue

EM

(in

△ D A B

) and green

MF

(in

△ B D C

)

∑ Formula

EF = \frac{m \cdot a + n \cdot b}{m + n} (D E : E A = m : n)

Length of a line parallel to the bases

$a$: the bottom base $A B$ (the side where $E A$ is)
$b$: the top base $D C$ (the side where $D E$ is)
$m, n$: $D E : E A = m : n$ (the piece near the top is $m$ )

Why: in $△ D A B$ , $EM = \frac{D E}{D A} a = \frac{m}{m + n} a$ , and in $△ B D C$ , $MF = \frac{BF}{BC} b = \frac{n}{m + n} b$ (since $BF : FC = A E : E D = n : m$ gives $BF : BC = n : (m + n)$ ). Adding gives the formula. Notice: the line weighs more towards the base it is nearer — if $E$ is near the top ( $m$ small), $EF$ is close to $b$ .

Three special lines

1. The mid-segment (joining the midpoints of the non-parallel sides): putting $m = n$ gives $EF = \frac{a + b}{2}$ , the average of the two bases. (In each triangle $EM = \frac{a}{2}$ and $MF = \frac{b}{2}$ .)

◭ Diagram

Mid-segment:

\frac{a}{2}

(blue)

+ \frac{b}{2}

(green)

= \frac{a + b}{2}

2. The segment joining the midpoints of the diagonals: both midpoints lie on the mid-segment line. Out of the mid-segment, $\frac{b}{2}$ is left over on each end, so the middle piece is $\frac{a + b}{2} - \frac{b}{2} - \frac{b}{2} = \frac{a - b}{2}$ .

◭ Diagram

MN

is the mid-segment;

P, Q

are the diagonal midpoints:

PQ = \frac{a - b}{2}

(pink)

3. The line through the crossing point $O$ of the diagonals ( $PQ ∥ A B$ ): $△ O A B \sim △ OC D$ gives $A O : OC = a : b$ . This leads to $PO = OQ = \frac{ab}{a + b}$ and the whole $PQ = \frac{2 ab}{a + b}$ (the harmonic mean of the two bases).

◭ Diagram

△ O A B \sim △ OC D

(blue, green) and the line

PQ

through the crossing point

Line	Length	Spot it
any line, $D E : E A = m : n$	$\frac{ma + nb}{m + n}$	a ratio is given
midpoints of the legs	$\frac{a + b}{2}$	“joining the midpoints”
midpoints of the diagonals	$\frac{a - b}{2}$	“midpoints of the diagonals”
through the diagonals’ crossing	$\frac{2 ab}{a + b}$	“through the point of intersection of the diagonals”

Here: $a$ = the longer base, $b$ = the shorter base, all in the same unit.

Steps

First see which line is asked: one with a ratio, the mid-segment, the diagonal-midpoint segment or the crossing-point line.
Apply the formula. If you cannot recall it, draw the diagonal and use Thales in the two triangles.
In reverse questions (finding a base) solve the formula for the base.

Example 13 · Basic

In trapezium

A BC D

A B ∥ D C

A B = 40

cm and

D C = 24

cm.

E

is on

A D

and

F

BC

with

EF ∥ A B

and

D E : E A = 1 : 3

. Find

EF

Solution, step by step

1Draw the diagonal $D B$ meeting $EF$ at $M$ . Then $D E : D A = 1 : 4$ .
2In $△ D A B$ , $EM ∥ A B$ : $EM = \frac{1}{4} \times 40 = 10$ cm.
3In $△ B D C$ , $MF ∥ D C$ and $BF : BC = A E : A D = 3 : 4$ : $MF = \frac{3}{4} \times 24 = 18$ cm.
4 $EF = EM + MF = 10 + 18 = 28$ cm.

Answer $EF = 28$ cm

By the formula:

\frac{1 \times 40 + 3 \times 24}{1 + 3} = \frac{112}{4} = 28

. ✓

Example 14 · Standard

The parallel sides of a trapezium are 18 cm and 12 cm. (a) Find the segment joining the midpoints of the non-parallel sides. (b) In another trapezium this segment is 15 cm and the shorter parallel side is 11 cm; find the longer one.

Solution, step by step

1(a) Mid-segment $= \frac{a + b}{2} = \frac{18 + 12}{2} = 15$ cm.
2(b) $\frac{a + 11}{2} = 15$ ⟹ $a + 11 = 30$ .
3So the longer side is $a = 19$ cm.

Answer (a) 15 cm (b) 19 cm

Check (b):

\frac{19 + 11}{2} = 15

. ✓

Example 15 · Standard

The parallel sides of a trapezium are 34 cm and 20 cm. Find the segment joining the midpoints of its diagonals. If the segment were 5 cm and the longer parallel side 26 cm, find the shorter one.

Solution, step by step

1The diagonal-midpoint segment is $\frac{a - b}{2}$ with $a$ the longer and $b$ the shorter side.
2First part: $\frac{34 - 20}{2} = 7$ cm.
3Second part: $\frac{26 - b}{2} = 5$ ⟹ $26 - b = 10$ ⟹ $b = 16$ cm.

Answer 7 cm; $b = 16$ cm

Here we subtract; adding is the mid-segment formula.

Example 16 · Advanced

The parallel sides of a trapezium are 30 cm and 20 cm. Find the length of the line through the crossing point of the diagonals parallel to the bases, and the ratio in which the diagonals cut each other.

Solution, step by step

1The diagonals $A C$ and $B D$ meet at $O$ . $A B ∥ D C$ gives equal alternate angles, and the vertically opposite angles are equal, so $△ O A B \sim △ OC D$ .
2So $A O : OC = A B : D C = 30 : 20 = 3 : 2$ . The diagonals cut in the ratio $3 : 2$ .
3For the line $PQ$ the formula gives $\frac{2 ab}{a + b} = \frac{2 \times 30 \times 20}{50} = 24$ cm.
4Check: in $△ A D C$ , $PO ∥ D C$ and $A O : A C = 3 : 5$ , so $PO = \frac{3}{5} \times 20 = 12$ . Whole $PQ = 12 + 12 = 24$ . ✓

Answer $PQ = 24$ cm; diagonals in the ratio $3 : 2$

It is not the average (25)! The harmonic mean is always a little less than the average.

Example 17 · Advanced

In trapezium

A BC D

A B ∥ D C

and

A B = 40

cm.

EF ∥ A B

with

D E : E A = 2 : 3

and

EF = 34

cm. Find

D C

Solution, step by step

1Formula: $EF = \frac{m \cdot a + n \cdot b}{m + n}$ with $m = 2, n = 3, a = 40$ .
2 $34 = \frac{2 \times 40 + 3 b}{5}$ ⟹ $170 = 80 + 3 b$ .
3 $3 b = 90$ , so $b = D C = 30$ cm.

Answer $D C = 30$ cm

Check:

\frac{80 + 90}{5} = 34

. ✓

Try it yourself

◭ Diagram

BasicTry it yourself first

In trapezium

A BC D

A B ∥ D C

A B = 50

cm and

D C = 30

cm.

E

is on

A D

and

F

BC

with

EF ∥ A B

and

D E : E A = 2 : 3

. Find

EF

◭ Diagram

BasicTry it yourself first

The parallel sides of a trapezium are 28 cm and 16 cm. How long is the segment joining the midpoints of the non-parallel sides?

◭ Diagram

BasicTry it yourself first

The segment joining the midpoints of the non-parallel sides of a trapezium is 34 cm and the shorter parallel side is 26 cm. Find the longer parallel side.

StandardTry it yourself first

The parallel sides of a trapezium are 46 cm and 30 cm. Find the segment joining the midpoints of its two diagonals.

StandardTry it yourself first

The longer parallel side of a trapezium is 40 cm and the segment joining the midpoints of its diagonals is 8 cm. Find the shorter parallel side.

StandardTry it yourself first

The parallel sides of a trapezium are 12 cm and 6 cm. A line through the point where the diagonals meet, parallel to the bases, ends on the non-parallel sides. How long is it?

StandardTry it yourself first

The parallel sides of a trapezium are 15 cm and 10 cm. A line through the point where the diagonals meet, parallel to the bases, ends on the non-parallel sides. How long is it?

AdvancedTry it yourself first

The parallel sides of a trapezium differ by 6 cm and the segment joining the midpoints of the non-parallel sides is 27 cm. Find the longer parallel side.

AdvancedTry it yourself first

The mid-segment of a trapezium is 20 cm and its diagonal-midpoint segment is 4 cm. Find the longer parallel side.

Ratio chains: draw a parallel, mass points

Ratios along crossing lines

When lines cross and a ratio on another line is asked: draw a parallel, or use mass points.

The idea

In many questions $D$ is on $BC$ , $E$ is on $A D$ , and $BE$ produced meets $A C$ at $F$ . We want $A F : FC$ . The figure has no parallel line, so Thales does not apply directly. We make a parallel ourselves: draw $D M ∥ BF$ from $D$ ( $M$ on $A C$ ). Now two small triangles are ready for Thales.

◭ Diagram

B D : D C = 2 : 3

A E : E D = 3 : 4

. Drawing

D M ∥ BF

(pink dashed) gives

A F : FM = 3 : 4

and

FM : MC = 2 : 3

In $△ A D M$ , $EF ∥ D M$ (as $BF ∥ D M$ ). Thales: $A F : FM = A E : E D = 3 : 4$ .
In $△ BCF$ , $D M ∥ BF$ . Thales: $FM : MC = B D : D C = 2 : 3$ .
$FM$ appears in both but with 4 and 2 parts. Make $FM$ equal (LCM 4): multiply the second ratio $2 : 3$ by 2 to get $4 : 6$ .
So $A F : FM : MC = 3 : 4 : 6$ and $A F : FC = 3 : (4 + 6) = 3 : 10$ .

The fast way: mass points

Drawing a parallel always works but is slow. Its fast form is mass points: put weights on the corners so that every line balances like a see-saw. Rule: if $P$ is on $X Y$ with $XP : P Y = m : n$ then put weight $n$ on $X$ and weight $m$ on $Y$ , and $P$ carries $m + n$ (the heavier end sits nearer, so the weights swap).

◭ Diagram

The same triangle:

B = 9, C = 6

give

D = 15

; on

A D

A = 20

and

E = 35

. Then

A F : FC = C : A = 6 : 20 = 3 : 10

Step	What to do	Here
1. weights on corners	$B D : D C = 2 : 3$ ⟹ $B = 3, C = 2$ ; $D = 5$	$B = 3, C = 2, D = 5$
2. weight of $A$ on the other line	$A E : E D = 3 : 4$ ⟹ $A \times 3 = D \times 4$	$A = \frac{20}{3}$ ; multiply all by 3: $A = 20, B = 9, C = 6, D = 15$
3. the line you need	$A F : FC = C : A$	$6 : 20 = 3 : 10$

Here: $A, B, C, D, E$ = the weight placed at that point (a number). The method is best when the lines start from the corners and the points lie on the sides.

RememberIt works with lines from different corners too (

A D

and

CE

): first the line with

D

fixes

B

and

C

, then the line with

E

fixes

A

, and at

P

the balance gives the ratio. Keep all weights whole numbers (multiply through if a fraction appears).

Steps

Draw the figure and write the given ratios on the pieces.
Place weights: on a line whose two ends are known swap the weights; match the weights at the shared point (by multiplying).
On the line you need, take the reversed ratio of the weights at its two ends.
If you like, confirm by drawing the parallel.

Example 18 · Standard

△ A BC

D

is on

BC

with

B D : D C = 2 : 3

E

is on

A D

with

A E : E D = 3 : 4

BE

produced meets

A C

F

. If

A C = 52

cm find

A F

(by drawing a parallel).

Solution, step by step

1Draw $D M$ through $D$ parallel to $BF$ , with $M$ on $A C$ .
2In $△ A D M$ , $EF ∥ D M$ : $A F : FM = A E : E D = 3 : 4$ .
3In $△ BCF$ , $D M ∥ BF$ : $FM : MC = B D : D C = 2 : 3$ .
4To equalise $FM$ multiply the second ratio by 2: $FM : MC = 4 : 6$ . So $A F : FM : MC = 3 : 4 : 6$ .
5 $A C = A F + FM + MC = 3 + 4 + 6 = 13$ parts $= 52$ cm, so 1 part $= 4$ cm.
6 $A F = 3 \times 4 = 12$ cm.

Answer $A F = 12$ cm (and $A F : FC = 3 : 10$ )

When chaining, always make the shared piece (

FM

) equal in both ratios.

Example 19 · Standard

The same triangle:

B D : D C = 2 : 3

and

A E : E D = 3 : 4

. Find

A F : FC

by mass points.

Solution, step by step

1In $B D : D C = 2 : 3$ the weights swap: $B = 3, C = 2$ , so $D = 3 + 2 = 5$ .
2On line $A D$ , $A E : E D = 3 : 4$ : $A \times 3 = D \times 4 = 5 \times 4 = 20$ , so $A = \frac{20}{3}$ .
3To remove the fraction multiply every weight by 3: $A = 20, B = 9, C = 6, D = 15$ .
4 $F$ is on $A C$ , so $A F : FC = C : A = 6 : 20 = 3 : 10$ .

Answer $A F : FC = 3 : 10$

Both methods agree. With mass points no parallel needs to be drawn.

Example 20 · Basic

△ A BC

D

is the midpoint of

BC

and

E

is the midpoint of the median

A D

BE

produced meets

A C

F

. If

A C = 45

cm find

A F

Solution, step by step

1These are midpoints, so $B D : D C = 1 : 1$ and $A E : E D = 1 : 1$ . Weights: $B = 1, C = 1, D = 2$ .
2 $A E : E D = 1 : 1$ ⟹ $A = D = 2$ . (Now $E = A + D = 4$ .)
3 $A F : FC = C : A = 1 : 2$ , i.e. $A F = \frac{1}{3} A C = 15$ cm.

Answer $A F = 15$ cm (and $FC = 30$ )

Remember: the “midpoint of the median” line cuts the opposite side in

1 : 2

Example 21 · Advanced

△ A BC

D

is on

BC

with

B D : D C = 2 : 5

and

E

is on

A B

with

A E : EB = 3 : 1

A D

and

CE

meet at

P

. Find

A P : P D

and

CP : PE

Solution, step by step

1From $D$ : $B D : D C = 2 : 5$ ⟹ $B = 5, C = 2$ , $D = 7$ .
2From $E$ : $A E : EB = 3 : 1$ ⟹ $A : B = 1 : 3$ . $B = 5$ , so $A = \frac{5}{3}$ . Multiply all by 3: $A = 5, B = 15, C = 6, D = 21$ .
3The weight at $E$ is $A + B = 20$ .
4 $P$ on line $A D$ : $A P : P D = D : A = 21 : 5$ .
5 $P$ on line $CE$ : $CP : PE = E : C = 20 : 6 = 10 : 3$ .

Answer $A P : P D = 21 : 5$ and $CP : PE = 10 : 3$

With lines from different corners the weights must agree at all the points at once.

Example 22 · Standard

△ A BC

D

is on

A B

with

A D : D B = 2 : 3

D E ∥ BC

is drawn (

E

A C

) and then

EF ∥ D C

(

F

A B

). Find

A F : F D : D B

Solution, step by step

1Take $A B = 5 u$ , so $A D = 2 u$ and $D B = 3 u$ . $D E ∥ BC$ gives $A E : EC = A D : D B = 2 : 3$ .
2In $△ A D C$ , $EF ∥ D C$ : $A F : F D = A E : EC = 2 : 3$ .
3Split $A D = 2 u$ in $2 : 3$ : $A F = \frac{2}{5} \times 2 u = 0.8 u$ and $F D = 1.2 u$ .
4 $A F : F D : D B = 0.8 : 1.2 : 3$ ; multiplying by 5 gives $4 : 6 : 15$ .

Answer $A F : F D : D B = 4 : 6 : 15$

Check:

4 + 6 + 15 = 25

and

A D = 4 + 6 = 10

, so

A D : A B = 10 : 25 = 2 : 5

. ✓

Try it yourself

◭ Diagram

figure not to scale

BasicTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 1 : 1

E

is on

A D

with

A E : E D = 1 : 1

BE

produced meets

A C

F

. Find

A F : FC

◭ Diagram

figure not to scale

BasicTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 1 : 1

E

is on

A D

with

A E : E D = 2 : 1

BE

produced meets

A C

F

. Find

A F : FC

◭ Diagram

figure not to scale

StandardTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 1 : 2

E

is on

A D

with

A E : E D = 2 : 3

BE

produced meets

A C

F

. Find

A F : FC

◭ Diagram

figure not to scale

StandardTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 3 : 2

and

E

is on

A D

with

A E : E D = 1 : 2

BE

produced meets

A C

F

. If

A C = 52

cm, find

A F

◭ Diagram

figure not to scale

StandardTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 2 : 3

and

E

is on

A D

with

A E : E D = 2 : 1

BE

produced meets

A C

F

. If

A C = 36

cm, find

A F

StandardTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 1 : 2

and

E

is on

A B

with

A E : EB = 1 : 1

A D

and

CE

meet at

P

. Find

A P : P D

StandardTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 3 : 2

and

E

is on

A B

with

A E : EB = 2 : 3

A D

and

CE

meet at

P

. Find

A P : P D

AdvancedTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 2 : 3

and

E

is on

A B

with

A E : EB = 3 : 2

A D

and

CE

meet at

P

. Find

CP : PE

AdvancedTry it yourself first

△ A BC

D

is on

BC

with

B D : D C = 4 : 3

and

E

is on

A B

with

A E : EB = 1 : 2

A D

and

CE

meet at

P

. Find

CP : PE

StandardTry it yourself first

△ A BC

D

is on

A B

with

A D : D B = 1 : 2

D E ∥ BC

is drawn (

E

A C

) and then

EF ∥ D C

(

F

A B

). Find

A F : F D : D B

AdvancedTry it yourself first

△ A BC

D

is on

A B

with

A D : D B = 3 : 2

D E ∥ BC

is drawn (

E

A C

) and then

EF ∥ D C

(

F

A B

). Find

A F : F D : D B

The bow-tie and crossed ladders

Lines crossing between parallels

Two lines cross at a point and end on two parallel lines: the facing triangles are similar; the pole-and-rope formula $p = \frac{ab}{a + b}$ .

The idea

Two lines $FC$ and $B D$ cross at $E$ with $FB ∥ C D$ , and the figure looks like a bow-tie. The two wings, $△ EFB$ and $△ E D C$ , are similar: $∠ FEB = ∠ D EC$ (vertically opposite) and, from the parallels, $∠ EFB = ∠ EC D$ , $∠ EBF = ∠ E D C$ (alternate angles).

◭ Diagram

Pink dashed: turning

△ EFB

half a turn about

E

puts it inside

△ E D C

with

FB

as a line parallel to

D C

— the bow-tie is really Thales’ picture in disguise

∑ Formula

\frac{FB}{C D} = \frac{FE}{EC} = \frac{BE}{E D}

Bow-tie:

FB ∥ C D

$FB, C D$: the two parallel sides
$FE, EC$: pieces from the crossing point to $F$ and to $C$ (facing each other)
$BE, E D$: pieces from the crossing point to $B$ and to $D$

Watch outMatching rule: the pieces going in opposite directions from the crossing point correspond:

FE \leftrightarrow EC

and

BE \leftrightarrow E D

(the two pieces of one line). Never pair

FE

with

E D

Crossed ladders / poles and ropes

Two vertical poles of heights $a$ and $b$ . A rope goes from the top of each to the foot of the other. The height $p$ at which the ropes cross does not depend on the distance $w$ between the poles. Why? Let $x$ be the distance from the left foot to the point below the crossing. By similar triangles $\frac{x}{w} = \frac{p}{a}$ and $\frac{w - x}{w} = \frac{p}{b}$ . Adding gives $1 = \frac{p}{a} + \frac{p}{b}$ .

◭ Diagram

Poles

a

(blue) and

b

(green); the ropes cross at

O

;

O H = p

∑ Formula

p = \frac{ab}{a + b} ⟺ \frac{1}{p} = \frac{1}{a} + \frac{1}{b}

Crossed ladders

$p$: height of the crossing above the ground
$a, b$: the heights of the two upright things (poles, walls)

This is also why the line through the crossing point of a trapezium’s diagonals is $\frac{2 ab}{a + b}$ : its half is $\frac{ab}{a + b}$ .

Steps

Check that two lines cross and run to two parallel lines (bow-tie) or that it is a poles-and-ropes situation.
For a bow-tie: pair the facing pieces, $\frac{FB}{C D} = \frac{FE}{EC} = \frac{BE}{E D}$ ; if a whole line is given share it in parts.
For poles and ropes: use $p = \frac{ab}{a + b}$ at once (ignore the distance).

Example 23 · Basic

FB ∥ C D

and

FC

B D

cross at

E

FB = 9

cm,

C D = 15

cm and

EC = 20

cm. Find

FE

and

FC

Solution, step by step

1Since $FB ∥ C D$ , $∠ EFB = ∠ EC D$ (alternate) and $∠ FEB = ∠ CE D$ (vertically opposite): $△ EFB \sim △ EC D$ ( $F \to C, B \to D$ ).
2Matching: $FB \leftrightarrow C D$ and $FE \leftrightarrow CE$ . $\frac{FE}{EC} = \frac{FB}{C D} = \frac{9}{15} = \frac{3}{5}$ .
3 $FE = \frac{3}{5} \times 20 = 12$ cm.
4The whole $FC = FE + EC = 12 + 20 = 32$ cm.

Answer $FE = 12$ cm, $FC = 32$ cm

The ratio is

3 : 5

, and

12 : 20

3 : 5

too. ✓

Example 24 · Standard

In the same bow-tie take

FB = 12

cm,

C D = 20

cm and

B D = 40

cm. Find

BE

and

E D

Solution, step by step

1Bow-tie ratio: $\frac{BE}{E D} = \frac{FB}{C D} = \frac{12}{20} = \frac{3}{5}$ .
2 $B D = BE + E D$ is $3 + 5 = 8$ parts $= 40$ cm, so 1 part $= 5$ cm.
3 $BE = 3 \times 5 = 15$ cm and $E D = 5 \times 5 = 25$ cm.

Answer $BE = 15$ cm, $E D = 25$ cm

BE

is on the side of the shorter parallel

FB

, so it is the smaller piece.

Example 25 · Standard

Two poles are 12 m and 20 m tall. A rope runs from the top of each to the foot of the other. At what height do the ropes cross? What if the poles are moved nearer or farther apart?

Solution, step by step

1Directly: $p = \frac{ab}{a + b} = \frac{12 \times 20}{12 + 20} = \frac{240}{32} = 7.5$ m.
2The distance $w$ does not appear in the formula. So moving the poles nearer or farther leaves the crossing at 7.5 m.
3Check: $\frac{1}{12} + \frac{1}{20} = \frac{5 + 3}{60} = \frac{8}{60} = \frac{1}{7.5}$ . ✓

Answer 7.5 m; changing the distance makes no difference

The crossing is always lower than both heights.

Example 26 · Advanced

In trapezium

A BC D

A B ∥ D C

A B = 15

cm,

D C = 10

cm and the diagonal

B D = 30

cm. The diagonals meet at

O

. Find

OB

O D

and the half

PO

of the line through

O

parallel to

A B

Solution, step by step

1 $A B ∥ D C$ makes $△ O A B$ and $△ OC D$ a bow-tie: $\frac{OB}{O D} = \frac{A B}{D C} = \frac{15}{10} = \frac{3}{2}$ .
2 $B D = OB + O D$ is $3 + 2 = 5$ parts $= 30$ cm, 1 part $= 6$ cm. $OB = 18$ cm and $O D = 12$ cm.
3In $△ D A B$ , $PO ∥ A B$ and $D O : D B = 12 : 30 = 2 : 5$ : $PO = \frac{2}{5} \times 15 = 6$ cm.
4Formula check: $\frac{ab}{a + b} = \frac{15 \times 10}{25} = 6$ . ✓

Answer $OB = 18$ cm, $O D = 12$ cm, $PO = 6$ cm

In a trapezium the diagonals’ crossing is a bow-tie too: the pieces are in the ratio of the parallel sides.

Try it yourself

◭ Diagram

BasicTry it yourself first

FB ∥ C D

and

FC

B D

cross at

E

. If

FB = 6

cm,

C D = 10

cm and

EC = 15

cm, find

FE

◭ Diagram

BasicTry it yourself first

FB ∥ C D

and

FC

B D

cross at

E

. If

FB = 8

cm,

C D = 12

cm and

EC = 18

cm, find

FE

◭ Diagram

StandardTry it yourself first

FB ∥ C D

and

FC

B D

cross at

E

. If

FB = 7

cm,

C D = 5

cm and

B D = 36

cm, find

BE

◭ Diagram

StandardTry it yourself first

FB ∥ C D

and

FC

B D

cross at

E

. If

FB = 9

cm,

C D = 15

cm and

B D = 48

cm, find

BE

◭ Diagram

BasicTry it yourself first

Two poles 10 m and 15 m tall stand 6 m apart. A rope is tied from the top of each pole to the foot of the other. At what height do the ropes cross?

◭ Diagram

StandardTry it yourself first

One pole is 6 m tall. With ropes tied crosswise from the top of each pole to the foot of the other, they cross at 3.6 m. How tall is the other pole?

◭ Diagram

StandardTry it yourself first

Two poles 14 m and 21 m tall stand 9 m apart. A rope is tied from the top of each pole to the foot of the other. At what height do the ropes cross?

StandardTry it yourself first

In trapezium

A BC D

A B ∥ D C

A B = 24

cm and

D C = 16

cm. The diagonal

A C = 50

cm and the diagonals meet at

O

. Find

OC

AdvancedTry it yourself first

Two poles are 6 m and 3 m tall. They were first 4 m apart and then moved 8 m apart (each time with ropes from the top of each to the foot of the other). At what height do the ropes cross the second time?

AdvancedTry it yourself first

The parallel sides of a trapezium are 36 cm and 12 cm. The line through the crossing point

O

of the diagonals, parallel to

A B

, meets

A D

P

. Find

PO

Right triangles and trigonometry

Altitude on the hypotenuse · tan

The altitude on the hypotenuse gives three similar triangles; a tan value tells you angles are equal.

The altitude on the hypotenuse

Three similar triangles

The altitude on the hypotenuse of a right triangle makes two smaller triangles similar to the whole; this gives formulas such as $A B^{2} = B D \cdot BC$ .

The idea

In $△ A BC$ , $∠ A = 9 0^{\circ}$ and $A D ⊥ BC$ ( $BC$ the hypotenuse). The figure then holds three right triangles and all three are similar. Reason: $△ A BC$ and $△ D B A$ share angle $B$ and each has a right angle, so they are similar by AA. In the same way $△ A BC$ and $△ D A C$ share angle $C$ . The remaining angle follows: $∠ B A D = 9 0^{\circ} - ∠ B = ∠ C$ .

◭ Diagram

Top: the original with the altitude

A D

. Bottom: the three pulled apart (orange arc = angle

B

, pink = angle

C

)

Each pair of triangles now gives one formula (match sides opposite the equal angles):

∑ Formula

A B^{2} = B D \times BC, A C^{2} = D C \times BC, A D^{2} = B D \times D C, A D \times BC = A B \times A C

Altitude on the hypotenuse (

∠ A = 9 0^{\circ}

A D ⊥ BC

)

$A B, A C$: the two sides forming the right angle
$BC$: the hypotenuse
$A D$: the altitude on the hypotenuse
$B D, D C$: the two pieces of the hypotenuse

◭ Diagram

See

A D^{2} = B D \times D C

as areas: the square on

A D

has the same area as the rectangle with sides

B D

and

D C

Formula	From which pair	How to remember
$A B^{2} = B D \times BC$	$△ A BC \sim △ D B A$	side squared = its own piece × the whole hypotenuse
$A C^{2} = D C \times BC$	$△ A BC \sim △ D A C$	the same
$A D^{2} = B D \times D C$	$△ D B A \sim △ D A C$	altitude squared = product of the two pieces
$A D \times BC = A B \times A C$	the area in two ways	altitude × hypotenuse = leg × leg

Two more useful facts: (1) adding the first two gives $A B^{2} + A C^{2} = BC (B D + D C) = B C^{2}$ — a proof of Pythagoras! (2) dividing them gives $A B^{2} : A C^{2} = B D : D C$ , i.e. the ratio of the squares of the legs = the ratio of the pieces. And $\frac{1}{A D ^{2}} = \frac{1}{A B ^{2}} + \frac{1}{A C ^{2}}$ .

Watch outThe most common slip: writing

A B^{2} = B D \times D C

. $A B^{2}$ is $B D \times BC$ : its own piece

B D

times the whole hypotenuse

BC

. Only

A D^{2}

uses the two pieces.

When a perpendicular is dropped from a point on a leg to the hypotenuse

In many questions a perpendicular $D E$ is dropped from a point $D$ on the leg $BC$ to the hypotenuse $A C$ . Then $△ CE D$ and $△ CB A$ share angle $C$ and $∠ E = ∠ B = 9 0^{\circ}$ , so $△ CE D \sim △ CB A$ ( $C \to C, E \to B, D \to A$ ). No formula to memorise, just similarity.

Steps

Spot the right angle and the perpendicular; decide whether it falls on the hypotenuse (formulas) or elsewhere (plain similarity).
Write the pieces $B D$ , $D C$ and the whole $BC = B D + D C$ separately.
Pick the formula of the side asked: leg² = piece × hypotenuse, altitude² = piece × piece.
Check with Pythagoras at the end.

Example 27 · Basic

△ A BC

∠ A = 9 0^{\circ}

A B = 15

cm and

A C = 20

cm. If

A D ⊥ BC

, find

BC

A D

B D

and

D C

Solution, step by step

1Pythagoras: $BC = 1 5^{2} + 2 0^{2} = 625 = 25$ cm.
2Area two ways: $A D \times BC = A B \times A C$ , so $A D = \frac{15 \times 20}{25} = 12$ cm.
3From $A B^{2} = B D \times BC$ , $B D = \frac{225}{25} = 9$ cm.
4From $A C^{2} = D C \times BC$ , $D C = \frac{400}{25} = 16$ cm.

Answer $BC = 25$ , $A D = 12$ , $B D = 9$ , $D C = 16$ cm

Check:

B D + D C = 9 + 16 = 25

and

A D^{2} = 144 = 9 \times 16

. ✓

Example 28 · Standard

△ A BC

∠ A = 9 0^{\circ}

and

A D ⊥ BC

. If

B D = 4

cm and

D C = 9

cm, find

A D

A B

and

A C

Solution, step by step

1 $A D^{2} = B D \times D C = 4 \times 9 = 36$ , so $A D = 6$ cm.
2The hypotenuse is $BC = B D + D C = 13$ cm.
3 $A B^{2} = B D \times BC = 4 \times 13 = 52$ , so $A B = 52 = 213$ cm.
4 $A C^{2} = D C \times BC = 9 \times 13 = 117$ , so $A C = 313$ cm.
5Check: $A B^{2} + A C^{2} = 52 + 117 = 169 = 1 3^{2}$ . ✓

Answer $A D = 6$ cm, $A B = 213$ cm, $A C = 313$ cm

The square formulas work directly even when the sides are not whole numbers.

Example 29 · Standard

The legs of a right triangle are 9 cm and 12 cm. Find the length of the altitude on the hypotenuse and verify it with

\frac{1}{A D ^{2}} = \frac{1}{A B ^{2}} + \frac{1}{A C ^{2}}

Solution, step by step

1Hypotenuse $= 81 + 144 = 15$ cm.
2 $A D = \frac{A B \times A C}{BC} = \frac{9 \times 12}{15} = 7.2$ cm.
3Check: $\frac{1}{81} + \frac{1}{144} = \frac{144 + 81}{81 \times 144} = \frac{225}{11664}$ and $\frac{1}{7. 2 ^{2}} = \frac{1}{51.84} = \frac{225}{11664}$ . ✓

Answer 7.2 cm

The area form is fastest: altitude = leg × leg ÷ hypotenuse.

Example 30 · Advanced

△ A BC

∠ A = 9 0^{\circ}

A D ⊥ BC

and

A B : A C = 2 : 3

. Find

B D : D C

and, if

BC = 26

cm,

B D

Solution, step by step

1 $A B^{2} = B D \times BC$ and $A C^{2} = D C \times BC$ . Dividing, $\frac{A B ^{2}}{A C ^{2}} = \frac{B D}{D C}$ .
2 $\frac{A B ^{2}}{A C ^{2}} = (\frac{2}{3})^{2} = \frac{4}{9}$ , so $B D : D C = 4 : 9$ .
3 $BC = B D + D C$ is $4 + 9 = 13$ parts $= 26$ cm, 1 part $= 2$ cm; $B D = 4 \times 2 = 8$ cm (and $D C = 18$ ).

Answer $B D : D C = 4 : 9$ ; $B D = 8$ cm

When the ratio of the legs is given the ratio of the pieces is its square.

Example 31 · Advanced

△ A BC

∠ B = 9 0^{\circ}

A B = 8

cm,

BC = 6

cm.

D

is on

BC

with

C D = 4

cm and

D E ⊥ A C

(

E

on the hypotenuse

A C

). Find

D E

and

CE

Solution, step by step

1Hypotenuse $A C = 64 + 36 = 10$ cm.
2 $△ CE D$ and $△ CB A$ share angle $C$ and $∠ E = ∠ B = 9 0^{\circ}$ , so $△ CE D \sim △ CB A$ ( $C \to C, E \to B, D \to A$ ).
3Corresponding sides: $C D \leftrightarrow C A$ , $CE \leftrightarrow CB$ , $E D \leftrightarrow B A$ . Multiplier $= \frac{C D}{C A} = \frac{4}{10} = \frac{2}{5}$ .
4 $D E = \frac{2}{5} \times 8 = 3.2$ cm and $CE = \frac{2}{5} \times 6 = 2.4$ cm.
5Check: in $△ CE D$ , $2. 4^{2} + 3. 2^{2} = 5.76 + 10.24 = 16 = 4^{2}$ . ✓

Answer $D E = 3.2$ cm, $CE = 2.4$ cm

A shared angle plus two right angles = AA; get the multiplier from the pair of sides given.

Try it yourself

◭ Diagram

BasicTry it yourself first

△ A BC

∠ A = 9 0^{\circ}

and

A D ⊥ BC

. If

B D = 8

cm and

D C = 18

cm, find

A D

◭ Diagram

BasicTry it yourself first

△ A BC

∠ A = 9 0^{\circ}

and

A D ⊥ BC

. If

B D = 9

cm and

BC = 25

cm, find

A B

◭ Diagram

BasicTry it yourself first

The legs of a right triangle are 6 cm and 8 cm. Find the length of the altitude on the hypotenuse.

◭ Diagram

StandardTry it yourself first

△ A BC

∠ A = 9 0^{\circ}

and

A D ⊥ BC

. If

A D = 12

cm and

B D = 9

cm, find

D C

◭ Diagram

StandardTry it yourself first

△ A BC

∠ A = 9 0^{\circ}

A B = 12

cm,

A C = 16

cm and

A D ⊥ BC

. Find

D C

StandardTry it yourself first

△ A BC

∠ A = 9 0^{\circ}

A D ⊥ BC

and

A B : A C = 3 : 4

. Find

B D : D C

◭ Diagram

StandardTry it yourself first

△ A BC

∠ A = 9 0^{\circ}

and

A D ⊥ BC

. If

A D = 15

cm and

B D = 5

cm, find

D C

◭ Diagram

AdvancedTry it yourself first

△ A BC

∠ B = 9 0^{\circ}

A B = 8

cm and

BC = 6

cm.

D

is on

BC

with

C D = 4

cm, and

D E ⊥ A C

. Find

D E

AdvancedTry it yourself first

In a right triangle the altitude on the hypotenuse is 12 cm and one leg is 15 cm. Find the other leg.

Similarity from trigonometric data

Equal tan · ratio triangle · shared altitude

When a $sin, cos, tan$ value is given it makes angles equal, and scaling the ratio triangle gives the real sides.

The idea

In a right triangle $sin θ = \frac{opposite}{hypotenuse}$ , $cos θ = \frac{adjacent}{hypotenuse}$ , $tan θ = \frac{opposite}{adjacent}$ . They depend only on the angle, not on the size of the triangle, because all right triangles with the same angle are similar. So every trigonometric value is really a statement of similarity.

Three techniques follow:

Technique	When	What to do
1. The ratio triangle	a value such as $sin θ = \frac{3}{5}$ and one side are given	build the Pythagorean triple ( $3, 4, 5$ ) from the value and scale it
2. Equal tan ⟹ equal angles	two right triangles have equal $tan$	the angles are equal, so similar by AA
3. A shared altitude	$tan$ of two angles with the same altitude $A D$	$B D = \frac{A D}{t a n B}$ , $D C = \frac{A D}{t a n C}$

Technique 1: the ratio triangle

If $tan θ = \frac{5}{12}$ draw a right triangle with opposite $5$ and adjacent $12$ ; the hypotenuse is $13$ by Pythagoras. Any real triangle with this angle is a multiple of this $5 - 12 - 13$ . So one side gives the whole triangle.

Value	Ratio triangle (opposite, adjacent, hypotenuse)
$sin θ = \frac{3}{5}$	$3, 4, 5$
$tan θ = \frac{5}{12}$	$5, 12, 13$
$cos θ = \frac{15}{17}$	$8, 15, 17$
$sin θ = \frac{7}{25}$	$7, 24, 25$

Here: $θ$ = the acute angle whose value is given; “opposite” is the side facing $θ$ , “adjacent” the one touching $θ$ that is not the hypotenuse.

Technique 2: equal tan

◭ Diagram

tan C = \frac{A F}{FC} = \frac{D E}{EC}

. The same angle

C

is in both triangles, so

△ A FC \sim △ D EC

When two right triangles have equal ratios of two corresponding sides ( $\frac{A F}{FC} = \frac{D E}{EC}$ ) their $tan$ are equal, so the angles are equal, so they are similar (also by SAS). And conversely: if equal $tan$ values are given, treat the angles as equal.

Technique 3: a shared altitude

◭ Diagram

The altitude

A D

is shared:

tan B = \frac{A D}{B D}

and

tan C = \frac{A D}{D C}

, so

tan B : tan C = D C : B D

(reversed)

∑ Formula

B D = \frac{A D}{tan B}, D C = \frac{A D}{tan C}, tan B : tan C = D C : B D

One altitude, two angles

$A D$: the altitude from $A$ to the base $BC$
$B D, D C$: the distances from the foot of the altitude to $B$ and $C$
$tan B, tan C$: $tan$ of the angles $B$ and $C$

CheckThe angle with the larger

tan

is steeper, so its base distance is shorter: a bigger

tan B

means a smaller

B D

. This catches reversed-ratio slips.

Steps

From the given value make equal angles or a ratio triangle (look for a Pythagorean triple).
Match the given and asked sides with the corresponding sides of the ratio triangle.
Find the multiplier $k$ and apply it to the asked side.

Example 32 · Basic

In a right triangle

A BC

∠ B = 9 0^{\circ}

and

sin A = \frac{3}{5}

. If the hypotenuse

A C = 20

cm find

BC

and

A B

Solution, step by step

1 $sin A = \frac{BC}{A C} = \frac{3}{5}$ . The ratio triangle has opposite $3$ , adjacent $4$ (Pythagoras: $25 - 9$ ), hypotenuse $5$ .
2The hypotenuse $5 \to 20$ gives $k = 4$ .
3 $BC = 3 \times 4 = 12$ cm and $A B = 4 \times 4 = 16$ cm.

Answer $BC = 12$ cm, $A B = 16$ cm

Check:

1 2^{2} + 1 6^{2} = 400 = 2 0^{2}

. ✓

Example 33 · Standard

In right triangle

PQR

∠ Q = 9 0^{\circ}

and

tan P = \frac{5}{12}

. If the perimeter is 60 cm find all three sides.

Solution, step by step

1 $tan P = \frac{QR}{PQ} = \frac{5}{12}$ , ratio triangle $5, 12, 13$ (Pythagoras: $25 + 144 = 13$ ).
2Its perimeter is $5 + 12 + 13 = 30$ . The real perimeter is 60, so $k = 2$ .
3 $QR = 10$ cm, $PQ = 24$ cm, $PR = 26$ cm.

Answer 10, 24 and 26 cm

The perimeter gives the multiplier: in similar triangles the perimeter scales by it too.

Example 34 · Standard

△ A BC

A F ⊥ BC

with

F

BC

D

is on

A C

and

D E ⊥ BC

(

E

BC

). If

A F = 12

cm,

FC = 9

cm and

EC = 6

cm, find

D E

and

A D

Solution, step by step

1In the right triangles $A FC$ and $D EC$ angle $C$ is shared: $tan C = \frac{A F}{FC} = \frac{12}{9} = \frac{4}{3}$ .
2In $△ D EC$ also $tan C = \frac{D E}{EC} = \frac{4}{3}$ , so $D E = \frac{4}{3} \times 6 = 8$ cm.
3 $A C = 1 2^{2} + 9^{2} = 15$ and $D C = 8^{2} + 6^{2} = 10$ cm (multiplier $\frac{2}{3}$ ).
4 $A D = A C - D C = 15 - 10 = 5$ cm.

Answer $D E = 8$ cm, $A D = 5$ cm

\frac{EC}{FC} = \frac{6}{9} = \frac{2}{3}

is the same multiplier:

△ D EC \sim △ A FC

Example 35 · Advanced

△ A BC

A D ⊥ BC

(

D

BC

tan B = \frac{3}{4}

tan C = \frac{1}{2}

and

BC = 20

cm. Find

A D

B D

and

D C

Solution, step by step

1Let $A D = h$ . $tan B = \frac{h}{B D} = \frac{3}{4}$ ⟹ $B D = \frac{4 h}{3}$ .
2 $tan C = \frac{h}{D C} = \frac{1}{2}$ ⟹ $D C = 2 h$ .
3 $BC = B D + D C = \frac{4 h}{3} + 2 h = \frac{10 h}{3} = 20$ , so $h = 6$ .
4 $A D = 6$ cm, $B D = \frac{4 \times 6}{3} = 8$ cm, $D C = 2 \times 6 = 12$ cm.

Answer $A D = 6$ , $B D = 8$ , $D C = 12$ cm

Check:

B D : D C = 8 : 12 = 2 : 3

and

tan C : tan B = \frac{1}{2} : \frac{3}{4} = 2 : 3

. ✓