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

(N/A) Given: $AB$ and $CD$ are two equal chords of a circle with centre $O$ which,when produced,intersect at $P$.
To prove: $PB = PD$.
Construction: Draw $OR \perp AB$ and $OQ \perp CD$. Join $OP$.
Proof: Since $OR \perp AB$ and $OQ \perp CD$ from the centre $O$ of the circle,
Therefore,$R$ is the mid-point of $AB$ and $Q$ is the mid-point of $CD$.
[Since the perpendicular from the centre to a chord bisects the chord]
Since $AB = CD$ [Given],
Therefore,$\frac{1}{2} AB = \frac{1}{2} CD$.
Therefore,$AR = CQ$ and $RB = QD$ $...(1)$
Since $AB = CD$,therefore $OR = OQ$ $...(2)$
[Since equal chords are equidistant from the centre]
Now,in right-angled $\Delta ORP$ and $\Delta OQP$,we have:
$\angle ORP = \angle OQP$ [Each $90^{\circ}$]
Hypotenuse $OP = $ Hypotenuse $OP$ [Common side]
$OR = OQ$ [From $(2)$]
$\Delta ORP \cong \Delta OQP$ [By $R.H.S.$ congruence rule]
$RP = QP$ [By $CPCT$] $...(3)$
Now,subtracting $(1)$ from $(3)$:
$RP - RB = QP - QD$
$PB = PD$
Hence,proved.