
5. Let $A B C$ be a triangle, $A D$ an altitude and $A E$ a median. Assume $B, D, E, C$ lie in that order on the line $B C$. Suppose the incentre of triangle $A B E$ lies on $A D$ and the incentre of $A D C$ lies on $A E$. Find the angles of triangle $A B C$.


Solution: Since $A D \perp B E$ and the incentre of $\triangle A B E$ lies on $A D$, it follows that $A B E$ is isosceles. In particular $\angle B A D=\angle D A E=\alpha$, say. Since $A E$ is the bisector of $\angle D A C$, it follows that $\angle E A C=\angle D A E=\alpha$. Moreover, we have

$$
\frac{A D}{A C}=\frac{D E}{C E}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_1e203c71c5d0062e7560g-2.jpg?height=296&width=575&top_left_y=1321&top_left_x=1382)

Since $B E=E C=\frac{a}{2}$, we also have $D E=\frac{1}{2} B E=\frac{a}{4}$. Thus we get

$$
\frac{A D}{A C}=\frac{D E}{E C}=\frac{a / 4}{a / 2}=\frac{1}{2}
$$

Since $\triangle A D E$ is a right-angled triangle and $A D=A C / 2$, it follows that $\angle A C D=30^{\circ}$. Hence $\angle D A C=60^{\circ}$. Since $\angle D A C=2 \alpha$, we get $\alpha=30^{\circ}$. Now $\angle A=3 \alpha$ and hence $\angle A=90^{\circ}$. This gives $\angle B=60^{\circ}$.
