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

To do:

We have to prove that the angle in a segment shorter than a semicircle is greater than a right angle.

Solution:

Let in a segment $ACB$ shorter than a semicircle and an angle $ACB$ be inscribed in it.

Join $OA$ and $OB$.

Arc $ADB$ subtends $\angle AOB$ at the centre and $\angle ACB$ at the remaining part of the circle.

Therefore,

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

$\angle AOB > 180^o$             (Reflex angle)

Therefore,

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

This implies,

$\angle ACB > 90^o$.

Hence proved.

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Updated on: 10-Oct-2022

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