bluecarneal wrote: Points $X$ and $Y$ are chosen on the sides $AB$ and $BC$ of the triangle $\triangle ABC$. The segments $AY$ and $CX$ intersect at the point $Z$. Given that $AY = YC$ and $AB = ZC$, prove that the points $B$, $X$, $Z$, and $Y$ lie on the same circle. Hi all i have anice proof see that take point $D \in\overline{CB}$ suchthat $CD=AZ$ now $\Delta ACY$ isoscles $\Rightarrow$$ AZDC$ isoscles trapizum $\Rightarrow AD=CZ=AB$ $\Delta ADB$ isoscles $\measuredangle{ACB}=\measuredangle{ABC}=2\alpha - \theta$ but $\measuredangle{CZY}=2 \alpha-\theta $ foom congrunty of $\Delta AZC , \Delta CZY$ no we have: $\measuredangle{CZY}=\measuredangle{XBY} \Rightarrow BYZX$ Cyclic eqilateral

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Let $l$ be the perpendicular bisector of $AC$. Let $B'$ be the symmetric of $B$ wrt $l$. Since $A$ is the symmetric of $C$ triangles $ABY$ and $CB'Y$ are congruent, so $CB' = AB$ and $\angle CB'Y = \angle ABY$. Also, since $B,Y$ and $C$ are collinear then $B',Y$ and $A$ are also collinear. Hence we have $CB' =AB = CZ$ and therefore $\angle CZY= \angle CB'Y = \angle ABY$ and we are done.

Another way shall be : Take a point $D$ in the plane such that $ABCD$ is a parallelogram. This impleis $ZC = CD=AB$ AND ITS GIVEN THAT $AY=YC$ doing simple angle chase using isoceles property of triangle and paralllel lines we can prove that $C$ IS midpoint of arc$ZD$ in thecircumcircle of $ZAD$; This would imply $AZCD$ is cyclic forcing $BXZY$ also to be..