Question

Prove that the angle in a segment shorter than a semicircle is greater than a right angle.

Answer 1

Given:

In figure,

A segment $ACB$ shorter than a semicircle.

To prove: $\angle ACB>{90}^{\circ }$

Construction:

Join $OA$ and $OB$.Proof: Arc $ADB$ subtends $\angle AOB$ at the centre and $\angle ACB$ at the remaining part of the circle.

We know that, angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at any point on the remaining part of the circle.

$\therefore \angle ACB=\frac{1}{2}\angle AOB$

But $\angle AOB>{180}^{\circ }$

$\therefore \angle ACB>\frac{1}{2}\times {180}^{\circ }$

$\Rightarrow \angle ACB>{90}^{\circ }$

Therefore, the angle in a segment shorter than a semicircle is greater than a right angle.

Hence, proved.