navi_09220114 wrote: A cyclic quadrilateral $ABCD$ has diameter $AC$ with circumcircle $\omega$. Let $K$ be the foot of the perpendicular from $C$ to $BD$, and the tangent to $\omega$ at $A$ meets $BD$ at $T$. Let the line $AK$ meets $\omega$ at $X$ and choose a point $Y$ on line $AK$ such that $\angle TYA=90^{\circ}$. Prove that $AY=KX$. Proposed by Anzo Teh Zhao Yang Note that $\angle CAT = \angle CKT$, so $CKAT$ is cyclic and $\psi := \angle CTA = \angle CKX$. Also $\angle AYT = 90^\circ \Rightarrow \varphi:=\angle ATY =\angle CAX$. Note that \[ \frac{KX}{AT}=^? \frac{AY}{AT}=sin \ \varphi = \frac{CX}{AC},\]or, equivalently, we want to prove that \[ \frac{AC}{AT}=\frac{CX}{KX}, \]but both them are equal to $tan \ \psi$ and we are done. $\square$

We have that $(ACKT)$ is cyclic with diameter $CT$. As triangles $CTA$ and $CKX$ are similar and $CTK$ and $ATY$ are similar. Thus $$AY=CK\cdot \frac{AT}{CT}=AT\cdot\frac{CK}{CT}=KX$$

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Violence complex bash solution: (a bit longggggg) WLOG let $\omega$ be unit circle on Argand Plane, $A=-1,C=1$. Apply complex foot formula get $\boxed{K=\frac{1}{2}(B+D+1-BD)}$, $\boxed{Y=\frac{1}{2}(T+X+X\overline{T}-1)}$; apply complex interception formula on unit circle (use line $AA$ and $BD$) get $\boxed{T=\frac{B+D+2BD}{1-BD}}$; apply complex collinear criterion on unit circle give $\boxed{X=\frac{K+1}{\overline{K}+1}}$, we want to prove $|\overrightarrow{AY}|=|\overrightarrow{XK}|$. On one hand, \begin{align*} |\overrightarrow{AY}|=|Y+1|&=\bigg|\frac{1}{2}\bigg(\frac{B+D+2BD}{1-BD}+X+\frac{X(\frac{1}{B}+\frac{1}{D}+\frac{2}{BD})}{1-\frac{1}{BD}}+1\bigg)\bigg|\\&=\bigg|\frac{1}{2}\bigg(\frac{B+D+2BD}{1-BD}+\frac{B+D+3-BD}{\frac{1}{B}+\frac{1}{D}+3-\frac{1}{BD}}+\frac{\frac{B+D+3-BD}{\frac{1}{B}+\frac{1}{D}+3-\frac{1}{BD}}(\frac{1}{B}+\frac{1}{D}+\frac{2}{BD})}{1-\frac{1}{BD}}+1\bigg)\bigg|\\&=\bigg|\frac{1}{2}\bigg(\frac{B+D+2BD}{1-BD}+\frac{B^2D+BD^2+3BD-(BD)^2}{B+D+3BD-1}+\frac{(B^2D+BD^2+3BD-(BD)^2)(B+D+2)}{(B+D+3BD-1)(BD-1)}+1\bigg)\bigg|\\&=\bigg|\frac{1}{2}\bigg(\frac{(B^2D+BD^2+3BD-(BD)^2)(B+D+BD+1)-(B+D+BD+1)(B+D+3BD-1)}{(B+D+3BD-1)(BD-1)}\bigg)\bigg|\\&=\bigg|\frac{1}{2}\bigg(\frac{(B^2D+BD^2-(BD)^2-B-D+1)(B+D+BD+1)}{(B+D+3BD-1)(BD-1)}\bigg)\bigg|\\&=\bigg|\frac{1}{2}\bigg(\frac{(B^2+D^2-(BD)^2-1)(BD-1)}{(B+D+3BD-1)(BD-1)}\bigg)\bigg|\\&=\bigg|-\frac{1}{2}\bigg(\frac{(BD)^2-B^2-D^2+1)}{B+D+3BD-1}\bigg)\bigg|\\ \end{align*}On the other hand, \begin{align*} |\overrightarrow{XK}|=|X-K|=\bigg|\frac{K+1}{\overline{K}+1}-K\bigg|&=\bigg|\frac{1-K\overline{K}}{\overline{K}+1}\bigg|\\&=\bigg|\frac{1-\frac{1}{4}(B+D+1-BD)(\frac{1}{B}+\frac{1}{D}+1-\frac{1}{BD})}{\frac{1}{2}(\frac{1}{B}+\frac{1}{D}+1-\frac{1}{BD})+1}\bigg|\\&=\bigg|\frac{BD-\frac{1}{4}(B+D+1-BD)(B+D+BD-1)}{BD+\frac{1}{2}(B+D+BD-1)}\bigg|\\&=\bigg|\frac{1}{2}\bigg(\frac{4BD-(B+D+1-BD)(B+D+BD-1)}{B+D+3BD-1}\bigg)\bigg|\\&=\bigg|\frac{1}{2}\bigg(\frac{(BD)^2-B^2-D^2+1)}{B+D+3BD-1}\bigg)\bigg|\\ \end{align*} As desire.