Question

Let ABC be a triangle and let $B{B}_1,C{C}_1$ be, respectively, the bisectors of $\angle B,\angle C$ with $B_1$ on AC and $C_1$ on AB. Let $E,F$ be the feet of perpendiculars drawn from $A$ onto $B{B}_1,C{C}_1,$ respectively. Suppose $D$ is the point at which the incircle of $\Delta ABC$ touches $AB$. Prove that $AD=EF$.

Answer 1

Observe that $\angle ADI=\angle AFI=\angle AEI={90}^{\circ }$
Hence A, F,D, I,E all lie on the circle with AI as diameter
We also know
$\angle BIC={90}^{\circ }+\angle A2=\angle FIE$
This gives
$\angle FAE={180}^{\circ }-({90}^{\circ }+\frac{\angle A}{2})$
$={90}^{\circ }-\frac{\angle A}{2}$
We also have $\angle AID={90}^{\circ }-\frac{\angle A}{2}$ . Thus $\angle FAE=\angle AID$
This shows the chords FE and AD subtend equal angles at the circumference of the same circle
Hence they have equal lengths, i.e., $FE=AD$