Question

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

Answer 1

Given: $AB$ and $CD$ two equal chords of a circle $O$ when produce intersect at $P$.

To prove: $PB=PD$

Construction: Draw $OR\perp AB$ and $OQ\perp CD$. Joint $OP$

Proof: $OR\perp AB$ and $OQ\perp CD$. rorm the centre $O$ od circle.

Refer image,

$\therefore R$ is mid-point of $AB$ and $Q$ is the mid point of $CD$

$(\because \perp$ from the centre to a chord bisects the chord)

$\because AB=CD$ [Given]

$\therefore \frac{1}{2}AB=\frac{1}{2}CD$

$\therefore AR=CQ$ and $RB=QD$...........(1)

$\therefore AR=CD$ and $\therefore RB=OQ$...........(2)

($\because$ Equal chords are equidistant from the centre)

Now, in right angled $\Delta sORP$ and $OPQ$. we have

$\angle ORP=\angle OQP$ [Each ${90}^0$]

hyp. $OP=hyp.OP$ [Common side]

$OR=OQ$ [From (2)]

$\therefore ORP\cong OQP$ [By R.H.S congruence]

$\therefore RP=QP$ [C.P.C.T]..........(3)

Now, subsracting (1) ffrom (3)

$RP-RB=QP-QD$

$\Rightarrow PB=PD$

Hence proved.