Question

Prove that the equal chords of a circle are equidistant from the center.

Answer 1

Given a circle with centre O and chords AB = cd
Draw $OP\perp AB$ and $OQ\perp CD$
Hence $AP=BP=\left(\frac{1}{2}\right)AB$ and CQ = QD = (\frac{1}{2}CD)\)
Also $\angle OPA={90}^{\circ }and\angle OQC={90}^{\circ }$
Since AB = CD
$\Rightarrow \left(\frac{1}{2}\right)AB=\frac{1}{2}CD$
$\Rightarrow AP=CQ$
In $\Delta$'s OPA and OQC
$\angle OPA=\angle OQC={90}^{\circ }$
$AP=CQ\left(proved\right)$
$OA=OC\left(Radii\right)$
$\therefore \Delta OPA\cong \Delta OQC$ (By RHS congruence criterion)
Hence OP = OQ (CPCT)