Review question

# If these centres lie on a line, why is the triangle isosceles? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8917

## Solution

Prove that the internal bisector of an angle of a triangle divides the opposite side in the ratio of the lengths of the containing sides.

All we have to do is prove the statement

$$$\label{eq:1} \frac{b}{c}=\frac{a_1}{a_2}.$$$

First we can apply the sine rule to the two smaller triangles created by the line $d$, giving

\begin{align*} I:&& \frac{d}{\sin \gamma} &= \frac{a_1}{\sin \frac{\alpha}{2}},&&\quad\\ II:&& \frac{d}{\sin \beta} &= \frac{a_2}{\sin \frac{\alpha}{2}}.\\ \end{align*}

Dividing $I$ by $II$ leads to

$$$\label{eq:2} \frac{\sin \beta}{\sin \gamma} = \frac{a_1}{a_2}.$$$

Our last step is to apply the sine rule on the big triangle $ABC$, giving

$$$\label{eq:3} \frac{\sin \beta}{\sin \gamma} = \frac{b}{c}.$$$

Combining the equations $\eqref{eq:2}$ and $\eqref{eq:3}$ finally proves $\eqref{eq:1}$.

In a triangle $ABC$, $I$ is the incentre, $H$ the orthocentre, and $O$ the circumcentre. Prove that $AI$ bisects $\angle OAH$.

Let’s sketch a triangle $ABC$ and construct the points $I$, $H$ and $O$ in it. This gives

To simplify the picture, we’ll leave out all lines not connected to $A$. We have to show that the two angles $\angle HAI$ and $\angle IAO$ are equal. Since we know that $AI$ is bisecting the angle in $A$, this comes down to proving $\delta_1 = \delta_2$, for the two angles shown below.

We start by taking a look at a simplified picture of the triangle and its orthocentre.

We know that the sum of the angles of triangle $ABD$ has to equal $\pi$, so $$$\label{eq:4} \delta_1 = \frac{\pi}{2}-\gamma.$$$

To define $\delta_2$, we sketch the triangle and its circumcentre.

Since we know that the distance from the circumcentre to each of the corner points is equal, $ABO$ is an isosceles triangle and $\angle ABO$ equals $\delta_2$. Hence $\angle OBC = \beta - \delta_2 = \angle OCB$, which leads to $\angle CAO = \gamma - \beta + \delta_2$.

Remembering that the full angle in $A$ is $\alpha$, and that we have $\alpha + \beta + \gamma = \pi$, we can write

\begin{align} \label{eq:5} \alpha &= \delta_2 + \gamma - \beta + \delta_2\\ \nonumber \alpha + \beta + \gamma &= 2 \delta_2 + 2\gamma\\ \nonumber \delta_2 &= \frac{\pi}{2} - \gamma. \end{align}

Comparing equation $\eqref{eq:4}$ with equation $\eqref{eq:5}$ proves $AI$ does indeed bisect $\angle OAH$.

Hence, or otherwise, prove that if $O$, $H$ and $I$ are collinear the triangle must be isosceles.

Suppose that $H, I$ and $O$ are collinear. Then $A, B$, and $C$ cannot all lie on one side of $OIH$, since $I$ is inside the triangle.

Suppose then that $A$ and $B$ lie on opposite sides of the line $OIH$. We have $OA = OB = r$, since $O$ is the circumcentre of the triangle $ABC$.

We have that $IA$ bisects the angle $OAH$, from our work above. Similarly $IB$ bisects the angle $OBH$.

From the diagram and our first result above, $\dfrac{x}{y} = \dfrac{r}{s} = \dfrac{r}{t}$, and so $s = t$.

This means that $AOH$ and $BOH$ are congruent, so $p = q$, and the line $OIH$ is a line of symmetry for the diagram.

This means that $OIH$ is the perpendicular bisector of $AB$, so $OIH$ is an altitude of the triangle, and $C$ lies on $OHI$. Thus $ABC$ is isosceles.