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Let $\alpha$ and $\omega$ be two circles such that $\omega$ passes through the center of $\alpha$. $\omega$ intersects $\alpha$ at points $A$ and $B$. Let $P$ be any point on the circumference of $\omega$. The lines $PA$ and $PB$ intersect $\alpha$ again at points $E$ and $F$, respectively. Prove that $AB = EF$. 


Source: BdMO 2019 National Secondary P5


Proof Based Problems  


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Solution

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Assume that the center of $\alpha$ is $O$. We have

$\angle FBE = \angle PBE = 180^\circ - \angle EPB - \angle PEB$

$=180^\circ-\angle APB-\angle PEB = \angle AOB-\angle PEB $

$=2\angle AEB - \angle AEB = \angle AEB$

So, on circle $ABEF$, $\angle AEB = \angle FBE \implies AB=EF$.

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