Question

In a circle, RS is the diameter and PQ chord with O as centre. Prove that angle RPO= angle OQR

Answer 1

Step-by-step explanation:

$\bf{\underline{\underline\red{Given:-}}}$

• RS is the diameter

• And PQ is the chord with centre O.

$\bf{\underline{\underline\blue{To\:Prove:-}}}$

• $\rm \angle RPO = \angle OQR$

$\bf{\underline{\underline\green{Proof:-}}}$

Let RS intersect PQ at M.

$\rm In \: \triangle PRM \: and \: \triangle QRM$

$\rm \implies \: RM \: = RM \: \: (common \: side)$

$\rm \implies \: \angle rmp \: = \angle rmq \: \:$

Line drawn from centre of the on the chord is perpendicular bisector of the chord ( It is perpendicular to chord) .So each angle is 90°.

$\rm \implies \: PM\: = MQ$

(RS is the perpendicular bisector of PQ)

$\rm \therefore \triangle PRM \cong \triangle QRM$

(By SAS congurency criterion)

Hence PM = QM.......(i). ( CPCTC- Corresponding parts of congruent triangle are congruent)

Now

In ∆RPO and ∆RQO

→ PO = QO ( Radius )

→ RO = RO ( Common side)

→ PR = QR [ From (i) ]

$\rm \therefore \triangle RPO \cong \triangle RQO$

( By SSS congurency criterion)

$\rm \therefore \angle PRM = \angle QRM \: \: \: ( By \: CPCTC)$

Hence, Proved