Angle inscribed in a semicircle is always a right angle.

**The proof is given below:**Let us assume a semicircle with diameter AB.

**Given:** ∠ACB is inscribed in semicircle.

**To Prove:** ∠ACB = 90°.

**Construction:** Join C and O.

**Proof:** Let ∠AOC = p and ∠BOC = q

∠p + ∠q = 180° (Linear pair angles)

AO = BO = CO (Radii)

Angles opposite to equal sides are equal.

∴ ∠CAO = ∠ACO = x(say)

and ∠CBO = ∠BCO = y(say)

In ΔAOC:

∠x + ∠x + ∠p = 180° (Sum of the angles of a triangle is 180°)

2∠x + ∠p = 180°.......

**1**In ΔBOC:

∠y + ∠y + ∠q = 180° (Sum of the angles of a triangle is 180°)

2∠y + ∠q = 180°.......

**2**Adding equations 1 and 2, we get

2(∠x + ∠y) + ∠p + ∠q = 360°

2(∠x + ∠y) + 180° = 360°

2(∠x + ∠y) = 180°

∠x + ∠y = 90°

∠ACB = 90°

Hence, angle inscribed in a semicircle is always a right angle.