# ANGLE SUBTENDED BY AN ARC AT THE CENTRE IS DOUBLE PROOF

## About "Angle subtended by an arc at the centre is double proof"

Angle subtended by an arc at the centre is double proof :

The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

 O is the centre of the circle. AXB is the arc. AOB is the angle  subtended by the arc AXB at the centre.

To prove : <AOB = 2 <ACB

Construction : Join CO and produce it to D

(ii) <OCA  =  <OAC

(angles opposite to equal sides are equal.)

(iii) In ΔAOC

<AOD  =  <OCA + <OAC

(Exterior angles of a triangle  =  Sum of the interior opposite angles)

(iv) <AOD  =  <OCA + <OCA

(substituting <OAC by <OCA)

(v) <AOD  =  2 <OCA (by addition)

(vi) Similarly in triangle BOC

<BOD  =  2 <OCB

(vii) <AOD + <BOD  =  2 <OCA + 2<OCB

=  2(<OCA + <OCB)

(<AOD + <BOD  =  <AOB and <OCA + <OCB  =  <ACB)

(viii) <AOB = 2 <ACB

Example 1 :

Find the value of x in the following figure.

Solution :

Using the theorem the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

(i) <AOB  =  2<ACB

<ACB  =  (1/2) <AOB

=  (1/2)  80°

=  40°

Example 2 :

Find the value of x in the following figure.

Solution :

reflex <AOB  =  2 <ACB

x = 2 ⋅ 100°  =  200°

Example 3 :

Find the value of x in the following figure.

<ABC + <BCA + <CAB  =  180°

56° + 90° + <CAB  =  180°

(a <BCA = angle on a semicircle  =  90°)

<CAB  =  180° -  146°

x  =  34°

Example 4 :

Find the value of x in the following figure.

OA = OB = OC ( radius )

<OCA  =  <OAC  =  25°

<OBC  =  <OCB = 20°

<ACB  =  <OCA + <OCB

= 25° + 20° =  45°

AOB  =  2 <ACB

x  =  2 (45°)

=  90°

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