Question

$PQ\text{and}RQ$ are chords of a circle equidistant from the centre. Prove that the diameter passing through $Q$ bisects $\angle PQR\text{and}\angle PSR$.

Answer 1

Step 1: Construct the figure.

Given: $PQ$ and $QR$ are two chords equidistant from the center

Diameter $QS$ bisects $\angle PQRand\angle PSR$.

Join $PSandRS$.

The figure of the given data is drawn as :

Step 2: Proving the required statement.

Since, chords are equidistant from the center are equal

$PQ=RQ$

In $△PQSand△RQS$

$PQ=RQ$

$\angle QPS=\angle QRS$ (each is at right angle as angle formed by a diameter is equal to 90)

$QS=QS$

∴By RHS Congruence Rule

$△PQS\cong △RQS$

By C.P.C.T.

$\angle PQS=\angle RQSand\angle PSQ=\angle RSQ$

$\therefore$$QS$ bisects $\angle PQRand\angle PSR$.

Hence, it is been proved that the diameter passes through $Q$ bisects $\angle PQR\text{and}\angle PSR$.