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P,Q \in AB, AC = AP,BC = BQ, R intersection of ... prove <ACB + < PRQ = 180^o,

Source: St Petersburg Olympiad 2016, Grade 9, P3

September 8, 2018

geometryperpendicular bisectorangle bisectorequal segments

Problem Statement

On the side

AB

of the non-isosceles triangle

ABC

, let the points

P

and

Q

be so that

AC = AP

and

BC = BQ

. The perpendicular bisector of the segment

PQ

intersects the angle bisector of the

\angle C

at the point

R

(inside the triangle). Prove that

\angle ACB + \angle PRQ = 180^o