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Mongolian Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it Mongolia

Let ABCDABCD be quadrilateral inscribed in the circle ω\omega. Simedian of the angle BB of the triangle △ABD\triangle ABD intersects ω\omega at point PP and simedian of the angle BB of the triangle △CBD\triangle CBD intersects ω\omega at point QQ. Prove that if CP∩AB=XCP \cap AB = X, AQ∩BC=YAQ \cap BC = Y then points XX, DD, YY lie on a line.

Solution

Let A′,D′,C′,P′,Q′,X′,Y′A', D', C', P', Q', X', Y' be images of points A,D,C,P,Q,X,YA, D, C, P, Q, X, Y transformed by inversion Iω(B,1)I_{\omega(B,1)} respectively.

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Figure 1

Figure 2

Then points A′,P′,D′,Q′,C′A', P', D', Q', C' are colinear and A′P′=P′D′A'P' = P'D', D′Q′=Q′C′D'Q' = Q'C'. This implies from △BA′D′∼△BAD\triangle BA'D' \sim \triangle BAD and fact that BPBP is simedian. Consequently X′=ω(BP′C′)∩A′BX' = \omega(BP'C') \cap A'B, Y′=ω(A′Q′B)∩BC′Y' = \omega(A'Q'B) \cap BC'. Now it is sufficient to prove that points B,X′,Y′,D′B, X', Y', D' are cyclic. Furthermore

∠X′A′P′=∠Q′Y′C′∠A′X′P′=∠Q′C′Y′⇒A′P′P′X′=Q′Y′Q′C′⇒P′D′P′X′=QY′D′Q′and∠X′P′D′=∠D′Q′Y′⇒△P′D′X′∼△Q′Y′D′⇒∠X′D′P′+∠Y′D′Q′=∠X′P′A′=180∘−∠X′BY′⇒∠X′BY′+∠X′D′Y′=180∘. \angle X'A'P' = \angle Q'Y'C' \quad \angle A'X'P' = \angle Q'C'Y' \Rightarrow \frac{A'P'}{P'X'} = \frac{Q'Y'}{Q'C'} \Rightarrow \frac{P'D'}{P'X'} = \frac{QY'}{D'Q'} \quad \text{and} \quad \angle X'P'D' = \angle D'Q'Y' \Rightarrow \triangle P'D'X' \sim \triangle Q'Y'D' \Rightarrow \angle X'D'P' + \angle Y'D'Q' = \angle X'P'A' = 180^\circ - \angle X'BY' \Rightarrow \angle X'BY' + \angle X'D'Y' = 180^\circ.
This completes the proof.

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