Let $AB$ be a chord in the circle $A_{1}$; $AE=EF=FB$; Point $X$ is on the circle, $PE$ meet with the circle at point $Y$, $PF$ meet with the circle at point $Z$. Show that $AY\cdot BZ=EF\cdot YZ$.
Proof: Apply the law of sine, we have
\[\frac{\sin{\angle AZY}}{\sin{\angle{YAZ}}}=\frac{AY}{YZ}=\frac{\sin{\angle AXY}}{\sin{\angle YXZ}}\]
\[=\frac{XF}{XA}=\frac{\sin\angle XAE}{\sin\angle XFA}=\frac{BF}{BZ}=\frac{EF}{BZ}\]
Comment: This problem is not kind of hard. It can be proved with pure geometry method easily too. Just consider the rotation of $\Delta AXZ$
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