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} $
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.
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