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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

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

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.