Question

Let $ABC$ be a triangle in which $\angle A={60}^o$. Let $BE$ and $CF$ be the bisectors of the angles $\angle B$ and $\angle C$ with $E$ on $AC$ and $F$ on $AB.$ Let $M$ be the reflection of $A$ in the line $EF.$ Prove that $M$ lies on $BC$.

Answer 1

Draw $AL\perp EF$ and extend it to meet $AB$ in $M$. We show that $AL=LM$.
First we show that $A,F,I,E$ are con-cyclic.
We have $\angle BIC={90}^o+\frac{\angle A}{2}={90}^o+{30}^o={120}^o$
$\therefore \angle FIE=\angle BIC={120}^o$.
Since $\angle A={60}^o$, it follows that $A,F,I,E$ are concyclic.
$\therefore \angle BEF=\angle IEF=\angle IAF=\frac{\angle A}{2}$.
This gives, $\angle AFE=\angle ABE+\angle BEF=\frac{\angle B}{2}+\frac{\angle A}{2}$.
Since $\angle ALF={90}^o$,
we see that $\angle FAM={90}^o\angle AFE={90}^o\frac{\angle B}{2}\frac{\angle A}{2}=\frac{\angle C}{2}=\angle FCM$.
$\Rightarrow F,M,C,A$ are concyclic.
It follows that $\angle FMA=\angle FCA=\frac{\angle C}{2}=\angle FAM$
Hence $FMA$ is an isosceles triangle.

But $FL\perp AM$.
Hence $L$ is the mid-point of $AM$ or $AL=LM$.