Question

In the figure $\left(i\right)$ given below, $P$ is the point of intersection of the chords $BC$ and $AQ$ such that $AB=AP$. Prove that $CP=CQ$

Answer 1

In ${\Delta }^u$ $APB$ & $CPQ$,

$\measuredangle B=\measuredangle Q$, (angle subtended by same chord on circumference)

$\measuredangle A=\measuredangle C$, (angle subtended by same chord $\left(BQ\right)$ on circumference)

$\measuredangle APB\measuredangle CPQ$ (V.O.A)

$\therefore$ $\Delta APB\sim \Delta CPQ$

$\therefore$ $\frac{AP}{AB}=\frac{CP}{CQ}$

$\Rightarrow \frac{CP}{CQ}=1$ ($\because$ $AP=AB$ (given))

$\therefore$ $CP=CQ$ Proved.