State the degree measure theorem and …

Ans. Degree Measure Theorem states that angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point on the circle.

Given : An arc PQ of a circle C with center O and radius r with a point R in arc $\stackrel\frown{QP}$ other than P or Q.To Prove : $\angle POQ = 2\angle PRQ$
Construction : Join RO and draw the ray ROM.

Proof : There will be three cases as

(i) $\stackrel\frown{PQ}$is a minor arc

(ii)$\stackrel\frown{PQ}$ is a semi-circle

(iii)$\stackrel\frown{PQ}$ is a major arc

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In each of these three cases, the exterior angle of a triangle is equal to the sum of two interior opposite angles.

Therefore, $\angle POM = \angle PRO + \angle RPO$  ............ (1) 

$\angle MOQ = \angle ORQ + \angle RQO$.......... (2)  

$In\space \triangle OPR \space and \space \triangle OQR$

Now, OP = OR and OR = OQ  (radii of the same circle)

$\angle PRO = \angle RPO$ and $\angle ORQ = \angle RQO$  (angles opposite to the equal sides are equal)

Hence, $\angle POM = 2\angle PRO$      .......... (3)

And $\angle MOQ = 2 \angle ORQ$     ............ (4)

Case (1) : adding equations (3) and (4) we get

$\angle POM + \angle MOQ = 2\angle PRO + 2\angle ORQ$

$\angle POQ = 2(\angle PRO + \angle ORQ) = 2 \angle PRQ$

$\angle POQ = 2\angle PRQ$

Hence Proved.

Similarly, you can proceed for case (ii) and case (iii)