Question

In Fig. 67, chords PQ and PS are equidistant from the centre O. R is a point such that RQ is perpendicular to PQ and RS is
perpendicular to PS. Prove that angle PRS and angle PRQ are equal.​

Answer 1

       $\underline{\bold{Given:}}$

       $PQ\text{ and }PS \text{ are equidistant from center }O$

       $\angle{PQR} = \angle{PSR} = 90^{o}$

       $\underline{\bold{To-Prove:}}$

       $\angle{PRS} = \angle{PRQ}$

       $\underline{\bold{Proof:}}$

       $\text{We know that, chords that are equidistant from the center are equal}$

$\implies PQ = PS$

       $\text{In }\triangle{PSR}\text{ and }\triangle{PQR}$

       $\angle{PSR} = \angle{PQR} = 90^{o} \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(\text{Given})$

       $PR = PR \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(\text{Common})$

       $PQ = PS \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(\text{Equidistant chords})$

$\implies \text{By RHS rule of congruency }\triangle{PSR} \cong \triangle{PQR}$

$\implies \angle{PRS} = \angle{PRQ} \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(\triangle{PSR}\cong\triangle{PQR})$

       $\text{Hence Proved}$