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Let P be the intersection of the convex quadrilateral ABCD. Let X,Y,Z be points on AB,BC,CD respectively such that $\frac{AX}{XB}=\frac{BY}{YC}=\frac{CZ}{ZD}=2$. Suppose that XY is tangent to the circumcircle of triangle CYZ and YZ is tangent to the circumcircle of triangle BXY. Prove that $\angle APD=\angle XYZ$

I did not make progress except for proving that $\Delta BXY\sim \Delta CYZ$, using alternate segment theorem.

Any help appreciated.

Thank you in advance.

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If you proved that $\Delta BXY\sim \Delta CYZ$ then $$\frac{BX}{CY}=\frac{BY}{CZ}$$ or $$\frac{\frac{1}{3}AB}{\frac{1}{3}BC}=\frac{\frac{2}{3}BC}{\frac{2}{3}DC}$$ or $$\frac{AB}{BC}=\frac{BC}{DC},$$ which says $$\Delta ABC\sim\Delta BCD.$$ Id est, $$\measuredangle APD=\measuredangle BDC+\measuredangle PCD=\measuredangle ACB+\measuredangle BCD-\measuredangle ACB=\measuredangle BCD=\measuredangle XYZ.$$

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  • $\begingroup$ Beautiful answer! I can't believe I did not see ABC is similar to BCD. $\endgroup$ – abc... Sep 15 '18 at 1:51

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