Question

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Answer 1

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 $\Delta OVT$ and $\Delta OUT$,
OV = OU (Equal chords of a circle are equidistant from the centre)
$\angle OVT=\angle OUT\left(Each{90}^{\circ }\right)$
OT = OT (Common)
$\therefore \Delta OVT\cong \Delta OUT$ (RHS congruence rule)
$\therefore VT=UT\left(ByCPCT\right)...\left(1\right)$

It is given that,
PQ = RS ... (2)
$\Rightarrow \frac{1}{2}PQ=\frac{1}{2}RS$
$\Rightarrow PV=RU...\left(3\right)$
On adding equations (1) and (3), we obtain
PV + VT = RU + UT
$\Rightarrow PT=RT...\left(4\right)$
On subtracting equation (4) from equation (2), we obtain
PQ - PT = RS - RT
$\Rightarrow QT=ST...\left(5\right)$
Equations (4) and (5) indicate that the corresponding segments of chords PQ and RS are congruent to each other.