Question

Chords $AB$ and $CD$ of a circle intersect each other at point $P$ such that $AP=CP$.

Show that : $AB=CD$.

Answer 1

When two chords intersect inside the circle, then,

$\Rightarrow AP\times PB=PD\times PC$

$\Rightarrow \frac{AP}{CP}=\frac{PD}{PB}$

But $AP=CP$ ...(i)

$\Rightarrow 1=\frac{PD}{PB}$

$\Rightarrow PD=PB$ ...(ii)

Adding (i) and (ii), we get,

$\Rightarrow AP+PD=CP+PB$

$\therefore AB=CD$