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
[Image Here ]
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)
Naveen Sharma 7 years, 7 months ago
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
[Image Here ]
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)
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