Question

prove degree measure theorem

Answer 1

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 QP⌢QP⌢ other than P or Q.To Prove : ∠POQ=2∠PRQ∠POQ=2∠PRQ
Construction : Join RO and draw the ray ROM.


Proof : There will be three cases as

(i) PQ⌢PQ⌢is a minor arc

(ii)PQ⌢PQ⌢ is a semi-circle

(iii)PQ⌢PQ⌢ is a major arc


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

Therefore, ∠POM=∠PRO+∠RPO∠POM=∠PRO+∠RPO ............ (1)

∠MOQ=∠ORQ+∠RQO∠MOQ=∠ORQ+∠RQO.......... (2)

In △OPR and △OQRIn △OPR and △OQR

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

∠PRO=∠RPO∠PRO=∠RPO and ∠ORQ=∠RQO∠ORQ=∠RQO (angles opposite to the equal sides are equal)

Hence, ∠POM=2∠PRO∠POM=2∠PRO .......... (3)

And ∠MOQ=2∠ORQ∠MOQ=2∠ORQ ............ (4)

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

∠POM+∠MOQ=2∠PRO+2∠ORQ∠POM+∠MOQ=2∠PRO+2∠ORQ

∠POQ=2(∠PRO+∠ORQ)=2∠PRQ∠POQ=2(∠PRO+∠ORQ)=2∠PRQ

∠POQ=2∠PRQ∠POQ=2∠PRQ

Hence Proved.

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

Answer 2

Answer:

Step-by-step explanation: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 QP⌢QP⌢ other than P or Q.To Prove : ∠POQ=2∠PRQ∠POQ=2∠PRQ

Construction : Join RO and draw the ray ROM.

Proof : There will be three cases as

(i) PQ⌢PQ⌢is a minor arc

(ii)PQ⌢PQ⌢ is a semi-circle

(iii)PQ⌢PQ⌢ is a major arc

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

Therefore, ∠POM=∠PRO+∠RPO∠POM=∠PRO+∠RPO ............ (1)

∠MOQ=∠ORQ+∠RQO∠MOQ=∠ORQ+∠RQO.......... (2)

In △OPR and △OQRIn △OPR and △OQR

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

∠PRO=∠RPO∠PRO=∠RPO and ∠ORQ=∠RQO∠ORQ=∠RQO (angles opposite to the equal sides are equal)

Hence, ∠POM=2∠PRO∠POM=2∠PRO .......... (3)

And ∠MOQ=2∠ORQ∠MOQ=2∠ORQ ............ (4)

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

∠POM+∠MOQ=2∠PRO+2∠ORQ∠POM+∠MOQ=2∠PRO+2∠ORQ

∠POQ=2(∠PRO+∠ORQ)=2∠PRQ∠POQ=2(∠PRO+∠ORQ)=2∠PRQ