Let $ABCD$ be a trapezoid with $AD \parallel BC$ and $\angle BCD < \angle ABC < 90^\circ$. Let $E$ be the intersection point of the diagonals $AC$ and $BD$. The circumcircle $\omega$ of $\triangle BEC$ intersects the segment $CD$ at $X$. The lines $AX$ and $BC$ intersect at $Y$, while the lines $BX$ and $AD$ intersect at $Z$. Prove that the line $EZ$ is tangent to $\omega$ iff the line $BE$ is tangent to the circumcircle of $\triangle BXY$.
Problem
Source: Macedonian National Olympiad 2021 P3
Tags: geometry, trapezoid, tangent, circles, cyclic quadrilateral, Parallel Lines, Angle Chasing
steppewolf
21.05.2021 01:58
Also my problem
mrleo_of_alf
21.05.2021 10:51
We have $\angle DBX = 180^{\circ} - \angle BYX = \angle CYX = \angle DAX$, so $DABX$ is a cyclic quadrilateral. Therefore, $\overline{ZA} \cdot \overline{ZD} = \overline{ZB} \cdot \overline{ZX}$, so $Z$ lies on the radical axis of $(AED)$ and $(BEC)$. It is well-known that $(AED)$ and $(BEC)$ are tangent to each other, so $Z$ must lie on the internal common tangent of those two circles, hence $ZE$ is tangent to $\omega$.
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MrOreoJuice
21.05.2021 13:06
Proof of $BE$ is tangent to the circumcircle of $\triangle BXY \implies EZ$ is tangent to $\omega$. $$\angle XEC = \angle XBC = \angle XZA \implies \ \text{XZAE is cyclic.}$$\begin{align*} \angle XEZ & = \angle XAZ \\ & = \angle XYB \\& = \angle XBE \end{align*}Hence $ZE$ is tangent to $(BEX) \equiv \omega$. Proof of $EZ$ is tangent to $\omega \implies BE$ is tangent to the circumcircle of $\triangle BXY$. $$\angle XEC = \angle XBC = \angle XZA \implies \ \text{XZAE is cyclic.}$$\begin{align*} \angle XBE & = \angle XEZ \\ &= \angle XAZ \\ &= \angle XYB \end{align*}Hence $BE$ is tangent to $(BXY)$.