Question

4) In the adjoining figure, the center of the
circle O is on the arm of an angle.
Observe the figure and show that: m(angleBAC)= 1/2 m(arcBXC)
(Hint - Join DC)
​

Answer 1

Answer:

$\boxed{\red{\sf\:m\:\angle\:BAC\:=\:\dfrac{1}{2}\:m\:(\:arc\:BXC\:)}}$

Step-by-step-explanation:

NOTE: Refer to the attachment for the diagram.

Given:

1) Point O is the centre of the circle.

2) $\sf\:\angle\:BAC$ is inscribed in an arc BAC.

3) $\sf\:\angle\:BAC$ intercepts are BXC of the circle.

To prove:

$\sf\:m\:\angle\:BAC\:=\:\dfrac{1}{2}\:m\:(\:arc\:BXC\:)$

Construction:

Join OC.

Proof:

We have given that, the point O is on the arm of the angle.

Now,

$\sf\:OA\:=\:OC\:\:\:-\:-\:[\:Radii\:of\:same\:circle\:]\\\\\\\therefore\sf\:\triangle\:OAC\:is\:an\:isosceles\:triangle\\\\\\\therefore\sf\:m\:\angle\:OAC\:=\:m\:\angle\:OCA\:\:\:-\:-\:[\:Isosceles\:triangle\:theorem\:]\\\\\\\sf\:Let\:,\\\\\\\sf\:m\:\angle\:OAC\:=\:m\:\angle\:OCA\:=\:x^{\circ}\\\\\\\therefore\sf\:m\:\angle\:BOC\:=\:m\:\angle\:OAC\:+\:m\:\angle\:OCA\:\:\:-\:-\:[\:Remote\:interior\:angles\:]\\\\\\\implies\sf\:m\:\angle\:BOC\:=\:x\:+\:x\\\\\\\implies\sf\:m\:\angle\:BOC\:=\:2x\\\\\\\implies\sf\:m\:\angle\:BOC\:=\:2\:m\:\angle\:BAC\\\\\\\implies\sf\:m\:\angle\:BAC\:=\:\dfrac{1}{2}\:m\:\angle\:BOC\\\\\\\sf\:Now\:,\\\\\\\sf\:m\:\angle\:BOC\:=\:m\:(\:arc\:BXC\:)\:\:\:-\:-\:[\:De finition\:of\:measure\:of\:minor\:arc\:]\\\\\\\implies\boxed{\red{\sf\:m\:\angle\:BAC\:=\:\dfrac{1}{2}\:m\:(\:arc\:BXC\:)}}$

Hence proved!