Angles adding up to 180 degrees

This is a Geometry theorem based on Angles adding up to 180 degrees. It is helpful for Mathematics Olympiad. Try to prove the statement!

Statement: Angles adding up to 180 degrees

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 circumcircle of BPC (prove this). Also extend AP to meet the circumcircle 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 $

Hence done.