Statement: ABC be an isosceles triangle with AB = AC. P be a point inside the triangle such that, \angle ABP = \angle BCP . Suppose M is the midpoint of BC. Show that \angle BPM + \angle APC = 180^o

Diagram 1

Diagram 1

Discussion:

Our first claim is, AB and AC are tangents to the circum circle of BPC (prove this). Also extend AP to meet the circum circle at G again. It is sufficient to show \angle GPC = \angle BPM .

Diagram 2

Diagram 2

Next we claim that IPCO and MPGO are cyclic (how?) .

Let \angle OGP = \angle OPG = y , \angle PCB = x

\angle OCB = \frac {\angle A}{2}

1. \angle IMP = y as MPGO is cyclic

2. \angle AIP = \frac {\angle A}{2} + x as IPCO is cyclic

So \angle BPM = \frac{\angle A}{2} + x - y

Also as OP = OC (radii), hence \angle OPC = \angle OCP = \frac{\angle A}{2} + x

Hence done.