Question

Two equal chords $AB$ and $CD$ of a circle when produced intersect at a point $P.$ Prove that $PB=PD.$

Answer 1

Given two equal chords $AB$ and $CD$ of a circle interesting at a point $P.$

To prove that $PB=PD.$

Construction : Join $OP,$ draw $OL$ $\perp$ $AB$ and $OM$ $\perp$ $CD.$

Proof: We have, $AB=CD$

OL = OM [equal chords are equidistant from the centre]

In $\Delta OLP$ and $\Delta OMP$, we have

OL = OM [proved above]

$\angle OLP=\angle OMP$ [each ${90}^{\circ }]$

And, $OP=OP$ [common side]

$\therefore \Delta OLP\cong \Delta OMP$ [by RHS congruence rule]

$LP=MP$ [by CPCT] ……(i)

Now, $AB=CD$

$\Rightarrow \frac{1}{2}\left(AB\right)=\frac{1}{2}\left(CD\right)$ [dividing both sides by 2]

$\Rightarrow$ $BL=DM$...............(ii)

[perpendicular draw from centre to the circle bisects the chord i.e., $AL=LB$ and $CM=MD$]

On subtracting Eq. (ii) from Eq. (i), we get

$LP–BL=MP–DM$

$\Rightarrow$ $PB=PD.$ hence proved.