# Theorem 5 - The Degree Measure of an Arc of a Circle, is Twice the Angle Subtended by it at any poin

# Theorem 5 - The Degree Measure of an Arc of a Circle, is Twice the Angle Subtended by it at any point of the Alternate Segment of the Circle with Respect to the Arc

**Theorem 5 : ** The degree measure of an arc of a circle is twice the angle subtended by it at any point of the alternate segment of the circle with respect to the arc.

**Given :** An arc PQ of a circle C(O, r) with a point R in arc other than P or Q.

**To Prove : ** âˆ POQ =2 âˆ PRQ

**Construction :** Join RO and draw the ray ROM.

**Proof : **There will be three cases as

(i) is a minor arc

(ii) is a semi-circle

(iii) 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, âˆ POM = âˆ PRO + âˆ RPO(i)

âˆ MOQ = âˆ ORQ + âˆ RQO(ii)

In Î”OPR and Î”OQR

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

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

Hence, âˆ POM = 2âˆ PRO(iii)

And âˆ MOQ = 2 âˆ ORQ (iv)

**Case (i) ** adding equations (iii) and (iv) we get

âˆ POM + âˆ MOQ = 2âˆ PRO + 2âˆ ORQ

âˆ´ âˆ POQ = 2(âˆ PRO + âˆ ORQ)

= 2 âˆ PRQ

âˆ´ âˆ POQ = 2âˆ PRQ

**Case (ii) ** âˆ POM + âˆ QOM = 180Â° = âˆ POQ

.^{.}. âˆ POQ = 2 âˆ ORP + 2âˆ ORQ

= 2 âˆ PRQ

**Case (iii)** âˆ POM + âˆ QOM

(180Â° - âˆ POR) + (180Â° - âˆ QOR)

= [(360Â° - (âˆ POR + âˆ QOR)]

= 360Â° - âˆ POQ

= âˆ POQ \âˆ POQ = 2 âˆ ORP + 2 âˆ ORQ = 2 âˆ PRQ

In the following figures find the angles x, y, z.

Refer to figure (i)

AB passes through centre O

âˆ´ AB is a diameter and âˆ ACB = z + y = 90Â° (Angle in a semi-circle)

In Î”ABC,

âˆ A + âˆ B + âˆ C = 180Â°

âˆ´ xÂ° + 55Â° + 90Â° = 180Â°

âˆ´ x = 35Â°

\ âˆ BDC = âˆ BCA = 35Â° (Angles in the same segment of the circles)

Now in Î”BCD, âˆ B + âˆ DCB + âˆ BDC = 180Â°

30^{o} + 55^{o} + y^{o} + 35^{o} = 180^{o}

âˆ´ y = 60^{o}

Now, âˆ ABD = âˆ ACD = 30Â° = z

âˆ´ x = 35Â°, y = 60Â° and z = 30Â°

Referring to figure (ii), we have 2xÂ° + 3xÂ° = 180Â° [opposite angles of a cyclic quad are supplementary).

.^{.}. 5x = 180Â°

âˆ´ x = 36Â°

Again 3yÂ° + 6yÂ° = 180Â°

âˆ´ 9y = 180Â°

âˆ´ y = 20Â°

Referring to figure (iii) we have

âˆ A + âˆ C = 180Â°

âˆ´ x + 2x = 180Â°

âˆ´ 3x = 180Â°

âˆ´ x = 60Â°

Now âˆ BOD = 2âˆ BAD [Angle at the centre is twice the angle in the alternate segment]

âˆ´z = 2x = 2Ã—60Â° = 120Â°