Geometry Incircle Problem

Question

Let $ABC$ be a triangle. Its incircle meets the sides $BC, \, CA$ and $AB$ in the points $D, \, E$ and $F$, respectively. Let $P$ denote the intersection point of $ED$ and the line perpendicular to $EF$ and passing through $F$, and similarly let $Q$ denote the intersection point of $EF$ and the line perpendicular to $ED$ and passing through $D$. Prove that $B$ is the mid-point of the segment $PQ$. as indicated in the

Answer 1

1. $\angle \, DEF = \frac{1}{2}(\angle \, A + \angle \, C) = 90^{\circ} - \frac{1}{2}\, \angle B \, $ because let's say triangles $AEF$ and $CED$ are isosceles;

2. $\angle \, DPF = \angle \, EPF = 90^{\circ} - \angle \, DEF = \frac{1}{2}\, \angle B$ because triangle $EFP$ is right-angled;

3. $\angle \, DPF = \frac{1}{2} \, \angle \, B = \frac{1}{2} \, \angle DBF$;

4. Draw circle $c_B$ centered at $B$ and of radius $BD = BF$ (tangents to incircle). Then $c_B$ passes through points $D$ and $F\,$;

5. Point $P$ lies on $c_B$ because $\angle \, DPF = \frac{1}{2} \, \angle DBF \, $;

6. Absolutely analogous arguments yield that point $Q$ lies on circle $c_B\,$;

7. $\angle \, PFQ = 90^{\circ}$ because by assumption line $FP$ is orthogonal to line $EF \equiv EQ$,

8. Therefore segment $PQ$ is a diameter of $c_B$ and since $B$ is the center of $c_B$, the point $B$ lies on $PQ$ and is its midpoint.