# BDMO 2016 Regional - Geometry

$$\triangle PQR \$$ is an isosceles triangle where $$PQ = PR.$$ $$X$$ is a point on the circumcircle of $$ΔPQR$$, such that it lies on the arc $$QR$$. The normal drawn from the point $$P$$ on $$XR$$ intersects $$XR$$ at point $$Y.$$ Prove that $$QX + YR = XY.$$

Reflect the $$R$$ across line $$PY$$ in to $$Z$$. So $$Z\in XR$$ and $$YZ = YR$$ and $$PZ = PR = PQ$$

Sp we have to prove that $$QX = XZ$$. Since $$\angle XZP = \pi-\angle RZP = \pi-\angle XRP = \angle XQP$$ and $$\angle QXP = \angle QRP = \angle RQP = \angle ZXP$$ we see that triangle $$XQP$$ is congruent to triangle $$XZP$$ (s.a.s.) and we are done.

• I think the reason of congruence is AAS instead.
– Mick
Dec 21 '18 at 2:24
• Could be also ASA @Mick
– Aqua
Dec 21 '18 at 8:06
• I mean at least not (s. a. s.).
– Mick
Dec 21 '18 at 17:32
• And why not @Mick
– Aqua
Dec 21 '18 at 17:59
• In your proof, you have mentioned 2 pairs of equal angles (and a common side). That means the reason to be quoted is either AAS or ASA.
– Mick
Dec 22 '18 at 8:51