To prove that $\angle B = \angle C$ or $\angle A$ is a right angle

Question

The question is:

In $\Delta ABC$, $M$ is the mid point of $BC$, $P$ is any point on $AM$, and $PE$ and $PF$ are perpendiculars drawn on to $AB$, $AC$ respectively. If $EF$ is parallel to $BC$ show that $\angle B = \angle C$ or angle A is $90^\circ$.

So I tried solving it via different methods. Let the intersection of line $EF$ and $AP$ be point $O$. I found out that $∆AEO \sim ∆ABM$ and $∆AOF \sim ∆AMC$ so $EO=OF$, i.e. $AO$ is the median for the parallel $EF$. I tried using Appolonius' theorem in $∆ AEF$, $∆PEF$ and $∆ABC$, but it's just getting messy. Any hint on this question would help.

Answer 1

$\angle AEP + \angle AFP = 90^\circ + 90^\circ = 180^\circ$ implies $A,E,P,F$ lie on a circle with diameter $AP$.

You have shown $EF$ is bisected at $O$. The perpendicular bisector of $EF$ passes through the center of the circle. Since the center lies on diameter $AP$,

• either $AP$ coincides with the perpendicular bisector of $EF$
• or $EF$ itself is another diameter.

The first case yields $\triangle AEF$ is isosceles hence is $\triangle ABC$. The second case implies $\triangle AEF$ hence $\triangle ABC$ is right angled at $A$.