Question

Question 11
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 $\Delta OLPand\Delta OMP$,
OL = OM [proved above]
$\angle OLP=\angle OMP\left[each{90}^{\circ }\right]$
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 [perpendicular draw from centre to the circle bisects the chord i.e., AL = LB and CM = MD]...............(ii)

On subtracting Eq. (ii) from Eq. (i), we get
LP – BL = MP – DM
$\Rightarrow$ PB = PD,hence proved.