Triangle Centres: Circumcentre, Centroid, Orthocentre and Incentre

Every triangle has four special points, and each is made by three lines meeting at one place (that the three always meet is the small miracle of geometry):

This chapter stands on the angle rules, isosceles triangles, Pythagoras and median lengths of the previous chapter (Plane Geometry). In each centre we first see why it exists (in the picture), then the rules, then the exam shortcut. The next chapter, Similar Triangles and Area Ratios, uses the $2:1$ of the centroid and the incentre.

The circumcentre from perpendicular bisectors, where it lies (acute, right, obtuse), the circumradius $R$ and $\angle BOC=2A$.

The idea

The perpendicular bisector of a side is the line that halves it and is perpendicular to it. Every point of that line is equally far from the two ends of the side. Take the perpendicular bisectors of $AB$ and $BC$; where they meet is the point $O$, and $OA=OB$ and $OB=OC$. So $OA=OC$ too, i.e. $O$ is also on the perpendicular bisector of $AC$: the three perpendicular bisectors meet at one point $O$, and $OA=OB=OC=R$. The circle with centre $O$ and radius $R$ passes through all three vertices: it is the circumcircle and $O$ the circumcentre.

Where does $O$ lie? In an acute triangle inside; in a right triangle at the midpoint of the hypotenuse; in an obtuse triangle outside, beyond the longest side (opposite the obtuse angle).

Why is $R=\frac{c}{2}$ in a right triangle? Complete the right triangle $ABC$ ($\angle A=90^{\circ }$) to the rectangle $ABDC$. Its diagonals $AD$ and $BC$ are equal and bisect each other, so their meeting point $M$ is the midpoint of $BC$ and $MA=MB=MC=\frac{BC}{2}$. So $M$ is the circumcentre.

Why $\angle BOC=2A$ (acute triangle): from $OA=OB$: $\angle OAB=\angle OBA=x$ and from $OA=OC$: $\angle OAC=\angle OCA=y$ (isosceles triangles). So $A=x+y$. In triangles $OAB$ and $OAC$ the angles at $O$ are $\angle AOB=180^{\circ }-2x$ and $\angle AOC=180^{\circ }-2y$, hence $\angle BOC=360^{\circ }-(180^{\circ }-2x)-(180^{\circ }-2y)=2(x+y)=2A$.

Steps

1. See the type of triangle (acute, right, obtuse) and decide where $O$ is.
2. For a length use $R=\frac{c}{2}$ (right) or $R=\frac{abc}{4\text{Area}}$; for an angle join $O$ to the vertices and use the isosceles triangles.
3. Check with results such as $\angle BOC=2A$ and $\angle OBC=90^{\circ }-A$.

The three medians meeting at one point, the $2:1$ rule and why it holds, the six equal parts and the condition $b^2+c^2=5a^2$ for perpendicular medians.

The idea

The line joining a vertex to the midpoint of the opposite side is a median. The three medians meet at one point $G$: the centroid (centre of gravity). A cardboard triangle balances on a pin at this point. The main rule: $G$ divides every median in the ratio $2:1$ from the vertex side, i.e. $AG:GD=2:1$.

Why $2:1$? Let the medians $BE$ and $CF$ meet at $G$. Take the midpoints $M$ and $N$ of $BG$ and $CG$. By the midpoint theorem $FE\parallel BC$, $FE=\frac{1}{2}BC$ in triangle $ABC$ and $MN\parallel BC$, $MN=\frac{1}{2}BC$ in triangle $GBC$. So $FE$ and $MN$ are parallel and equal: $FENM$ is a parallelogram and its diagonals bisect each other. Then $GE=GM=\frac{1}{2}BG$, i.e. $BG=2GE$ and $BG:GE=2:1$.

Perpendicular medians: if $BE\perp CF$ then $\angle BGC=90^{\circ }$, so $B{G}^2+C{G}^2=B{C}^2$. With $BG=\frac{2}{3}BE$, $CG=\frac{2}{3}CF$ and Apollonius ($B{E}^2=\frac{2a^2+2c^2-b^2}{4}$, $C{F}^2=\frac{2a^2+2b^2-c^2}{4}$) we get $\frac{4}{9}\cdot \frac{4a^2+b^2+c^2}{4}=a^2$, i.e. $b^2+c^2=5a^2$: $BC=\sqrt{\frac{A{B}^2+A{C}^2}{5}}$.

Two more useful things: (1) the three medians cut the triangle into six small triangles of equal area (the reason comes in the next chapter). So $△GBC$ has $\frac{1}{3}$ of the total area. (2) If $AG=BC$ then $GD=\frac{1}{3}AD=\frac{1}{2}BC=BD=DC$: in triangle $BGC$, $\angle DGB=\angle DBG=x$ and $\angle DGC=\angle DCG=y$, so $2x+2y=180^{\circ }$ and $\angle BGC=x+y=90^{\circ }$.

Steps

1. Spot which median it is and which part $G$ cuts off.
2. Take $\frac{2}{3}$ on the vertex side, $\frac{1}{3}$ on the side.
3. If the median length is unknown use Apollonius; for area questions use the $\frac{1}{3}$ and $\frac{1}{6}$ parts.

The three altitudes meeting, where $H$ lies in acute, right and obtuse triangles, $\angle BHC=180^{\circ }-A$, the orthocentric group of four points and the distance from $O$ to $H$.

The idea

The altitude from a vertex is the perpendicular from it to the opposite side (or its extension). The three altitudes meet at one point $H$: the orthocentre. This holds because drawing from each vertex a line parallel to the opposite side gives a bigger triangle in which the altitudes of $ABC$ are the perpendicular bisectors of the bigger triangle’s sides, and their meeting follows from the circumcentre argument.

Where is $H$? In an acute triangle inside. In a right triangle at the right-angle vertex (two sides are perpendicular to each other, so they are two of the altitudes and meet at that vertex). In an obtuse triangle outside, beyond the obtuse vertex.

Why $\angle BHC=180^{\circ }-A$? In an acute triangle take $BE\perp AC$ and $CF\perp AB$. In quadrilateral $AFHE$, $\angle AFH=\angle AEH=90^{\circ }$, so $\angle FHE=360^{\circ }-90^{\circ }-90^{\circ }-A=180^{\circ }-A$. Being vertically opposite, $\angle BHC=\angle FHE=180^{\circ }-A$.

The group of four points: in an acute triangle, for any three of the four points $A,B,C,H$ the fourth is the orthocentre of the triangle they make. (For example $A$ is the orthocentre of $△HBC$.) In an obtuse triangle $H$ is outside and the same works in reverse: the obtuse vertex is the orthocentre of the triangle made by $H$ and the other two vertices.

In a right triangle $OH=R$: $H$ is the right-angle vertex and $O$ the midpoint of the hypotenuse; their distance is the median, which equals $\frac{c}{2}=R$.

Steps

1. Spot the type of triangle and where $H$ is.
2. If an angle is asked, take the angle at that vertex and use $180^{\circ }-A$; for other angles (like $\angle HBC=90^{\circ }-C$) use the right triangle.
3. For a length use $R$ for $OH$ in the right case, otherwise the formula $9R^2-\sum a^2$.

The three angle bisectors meeting, the incircle and inradius $r$, $\angle BIC=90^{\circ }+\frac{A}{2}$, the excentre and the angle $\frac{|B-C|}{2}$ made by an external bisector.

The idea

Every point of an angle bisector is equally far from the two arms of the angle. Let the bisectors of $\angle B$ and $\angle C$ meet at $I$; then $I$ is equally far from $AB$ and $BC$ and also from $BC$ and $CA$, hence from $AB$ and $CA$ too, i.e. $I$ is also on the bisector of $\angle A$: the three angle bisectors meet at one point $I$, the incentre. Its distance from the three sides is the same, $r$ (the inradius), and the circle with this radius touches all three sides: the incircle.

The incentre is always inside the triangle (acute, right or obtuse). It is the only centre equally far from the three sides; the circumcentre is the one equally far from the three vertices.

Why $\angle BIC=90^{\circ }+\frac{A}{2}$? $\angle IBC=\frac{B}{2}$ and $\angle ICB=\frac{C}{2}$. In triangle $BIC$: $\angle BIC=180^{\circ }-\frac{B+C}{2}=180^{\circ }-\frac{180^{\circ }-A}{2}=90^{\circ }+\frac{A}{2}$.

The excentre: where the external angle bisectors at $B$ and $C$ meet is the point $I_a$ (on the opposite side of $BC$ from $A$). The external and internal bisectors are perpendicular, so $\angle IB{I}_a=90^{\circ }$ and $\angle B{I}_{a}C=90^{\circ }-\frac{A}{2}$. The circle about $I_a$ touches $BC$ and the extensions of $AB$, $AC$: the excircle.

Why $r=\frac{\text{Area}}{s}$? Join $I$ to the three vertices: the triangle splits into three triangles with height $r$ and bases $a,b,c$. Total area $=\frac{1}{2}r(a+b+c)=r\cdot s$. In a right triangle $r=\frac{a+b-c}{2}$ ($a,b$ the legs, $c$ the hypotenuse).

An external bisector and an angle: the external bisector of $A$ meets $CB$ produced at $M$; then $\angle AMB=\frac{|B-C|}{2}$. (Why: $\angle MAC=90^{\circ }+\frac{A}{2}$ and in triangle $AMC$, $\angle M=180^{\circ }-C-\left(90^{\circ }+\frac{A}{2}\right)=\frac{B-C}{2}$.)

Steps

1. Spot whether the bisectors are internal (incentre) or external (excentre).
2. For an angle: the $90^{\circ }\pm \frac{A}{2}$ formula; or take the halves of $B$ and $C$ and use the sum in triangle $BIC$.
3. For lengths: find the area (Heron or $\frac{1}{2}ab$), find $s$, then $r=\frac{\text{Area}}{s}$.

Comparing the four centres, where they lie in special triangles (equilateral, isosceles, right) and the Euler line.

The idea

The Euler line: in any triangle $O$, $G$ and $H$ lie on one straight line, with $G$ between them so that $OG:GH=1:2$, i.e. $H$ is three times as far from $O$ as $G$ is. (Only in the equilateral triangle are the three the same point.) A hint of the reason: $G$ divides a median $2:1$ and the midpoint $D$ of $BC$ is at distance $\frac{1}{2}AH$ from $O$, so triangles $AHG$ and $DOG$ give the ratio $2:1$. $I$ is generally not on this line.

Steps

1. Write down which centres the question has and see the type of triangle.
2. If the distance between two of $O,G,H$ is known, get the third from $OG:GH=1:2$.
3. Use the symmetry of special triangles (right, equilateral, isosceles) to place the centres.

A one-page table of the four centres, exam-style mixed questions and revision cards.