Question

Suppose two chords of a circle are equidistant from the centre of the circle. Prove that the chords have equal length.

Answer 1

In the figure, $O$ is the center of the circle. $AB$ and $CD$ are the chords of a circle which are equidistant from the center i.e $OP=OQ$.

To prove: $AB=CD$

Proof: In triangles $PBO$ and $QDO$

$\angle BPO=\angle DQO={90}^0$

$PO=QO$ (data)

$OB=OD$ (radil)

$PBO=QDO$ (RHS)

$PB=DQ$

Therefore, $AB=CD$

Hence, the chords have equal length.