Question

PQ and RQ are chords of a circle equidistant from the centre and S is a point on the circle,such that QS forms diameter. Prove that the diameter QS bisects $\angle$PQR and $\angle$PSR.

Answer 1

Given: PQ and RQ are two chords of a circle equidistant from the centre O and QS=Diameter

To prove: Diameter QS passing through Q bisects $\angle$PQR and $\angle$PSR

Proof:

Since, chords are equidistant from the centre are equal.

∴ PQ = RQ

In $△$PQS and $△$RQS,

PQ = RQ ( chords equidistant from centre are equal)

$\angle$QPS = $\angle$QRS (Each equal to ${90}^{\circ }$)

And QS=QS (Common)

$△$PQS$\cong$ $△$RQS (By RHS theorem of congruence)

$\angle$PQS = $\angle$RQS (C.P.C.T)

And $\angle$PSQ = $\angle$RSQ (C.P.C.T)
Hence proved.