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 \(\angle BPM + \angle APC = 180^o \)

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 \(\angle GPC = \angle BPM \).

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 \(\angle OPC \) = \(\angle OCP \) = \(\frac{\angle A}{2} + x \)

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