National problem · Geometry

Let \(ABC\) be an acute triangle with \(AB < AC\), and let \(D\) be the midpoint of its side \(BC\). Let \(p\) be the image of the line \(AD\) under reflection in the bisector of the angle \(BAC\), and let \(P\) be the foot of the perpendicular dropped from the vertex \(C\) to the line \(p\). Prove that

\angle APD = \angle BAC .

The median \(AD\), the bisector of \(\angle BAC\) (dotted), its reflection \(p\) of \(AD\), and the foot \(P\) of the perpendicular from \(C\). The foot need not lie inside the triangle.