If two equal chords of a circle intersect within the circle

Solution:

Let PQ and RS be two equal chords of a given circle and they are intersecting each other at point T.



Draw perpendiculars OV and OU on these chords.

In $\triangle O V T$ and $\Delta O U T$,

OV = OU (Equal chords of a circle are equidistant from the centre)

$\angle O V T=\angle O U T\left(\right.$ Each $\left.90^{\circ}\right)$

OT = OT (Common)

$\therefore \Delta$ OVT $\cong \triangle O U T(R H S$ congruence rule $)$

$\therefore \mathrm{VT}=\mathrm{UT}(\mathrm{By} \mathrm{CPCT}) \ldots(1)$

It is given that,

$P Q=R S \ldots(2)$

$\Rightarrow \frac{1}{2} \mathrm{PQ}=\frac{1}{2} \mathrm{RS}$

$\Rightarrow P V=R \cup \ldots(3)$

On subtracting equation (4) from equation (2), we obtain

$P Q-P T=R S-R T$

$\Rightarrow \mathrm{QT}=\mathrm{ST} \ldots(5)$

Equations (4) and (5) indicate that the corresponding segments of chords $P Q$ and $R S$ are congruent to each other.