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Geometry Difficulty 5.1 AIME, harder Prove it Belarus

Two circles ω1\omega_1 and ω2\omega_2 intersect each other at points XX and YY. Two lines pass through YY: one intersects ω1\omega_1 and ω2\omega_2 for the second time at points AA and BB respectively and another intersects ω1\omega_1 and ω2\omega_2 for the second time at points CC and DD respectively. The line ADAD intersects ω1\omega_1 and ω2\omega_2 for the second time at points PP and QQ respectively such that YP=YQYP = YQ.
Prove that circumcircles of △BCY\triangle BCY and △PQY\triangle PQY touch each other.
(Palina Chernikava)

Solution

The quadrilaterals ACYPACYP and BDQYBDQY are cyclic, so
∠ACD=∠ACY=∠YPQ and ∠ABD=∠YBD=∠YQP. \angle ACD = \angle ACY = \angle YPQ \text{ and } \angle ABD = \angle YBD = \angle YQP.
Since the triangle YPQYPQ is isosceles, ∠YPQ=∠YQP\angle YPQ = \angle YQP whence ∠ACD=∠ABD\angle ACD = \angle ABD. So the quadrilateral ACBDACBD is cyclic. Therefore
∠CYP=∠CYA+∠AYP=∠BYD+(∠YPQ−∠YAP)==∠YPQ+(∠BYD−∠BAD)=∠YQP+∠ADC==∠YQP+∠ABC=∠YQP+∠YBC. \begin{aligned} \angle CYP &= \angle CYA + \angle AYP = \angle BYD + (\angle YPQ - \angle YAP) = \\ &= \angle YPQ + (\angle BYD - \angle BAD) = \angle YQP + \angle ADC = \\ &= \angle YQP + \angle ABC = \angle YQP + \angle YBC. \end{aligned}
The last equality is equivalent to the tangency of the circumcircles of the triangles BCYBCY and PQYPQY.

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