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Singapore Mathematical Olympiad (SMO)

Geometry Difficulty 5.8 AIME, harder Prove it Singapore

In the triangle $ABC$ , $\angle B > 90^\circ$ , the incircle touches the sides $BC$ and $CA$ at $D$ and $E$ , respectively. The lines $ED$ and $AB$ intersect at $P$ . The incircle of the triangle $AEP$ touches the sides $PE$ and $AP$ at $D_1$ and $E_1$ , respectively. The lines $E_1D_1$ and $AE$ intersect at $P_1$ . Suppose $P, C, E, B$ are concyclic. Prove that $BE$ is parallel to $PP_1$ .

Solution

Let $\angle EPB = x$ . Since $P, C, E, B$ are concyclic, $x = \angle EPB = \angle BCA$ and therefore $\triangle AEP \sim \triangle ABC$ . Thus the points $(A, B, C, D, E, P)$ correspond to the points $(A, E, P, D_1, E_1, P_1)$ in 2 similar configurations associated with triangles $ABC, AEP$ . Therefore $\angle EPA = \angle E_1 P_1 A = x$ and $\angle CBE = \angle PEE_1$ .
Thus $P, P_1, E, E_1$ are concyclic and $E_1 P_1 \parallel BC$ and so $\angle PP_1 E_1 = \angle PEE_1 = \angle CBE$ , $\angle E_1 P_1 B = \angle CBP_1$ .

Therefore $\angle PP_1B = \angle PP_1E_1 - \angle E_1P_1B = \angle CBE - \angle CBP_1 = \angle P_1BE$ . Hence $BE \parallel PP_1$ .

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