Given: \(\angle\)ACB is any angle inscribed in a semicircle of a circle with center O. To Prove: \(\angle\)ACB is a right angle. Proof: Let \(\overset{\huge\frown}{APB}\) be the arc of the semicircle formed by diameter AB. The central angle subtended by this arc is \(\angle\)AOB, and the inscribed angle subtended by the same arc is \(\angle\)ACB. \[ \therefore \angle\text{AOB} = 2 \angle\text{ACB} \quad \text{——— (i)} \] Since AB is a diameter, \(\angle\)AOB is a straight angle. \[ \therefore \angle\text{AOB} = 180^\circ = 2 \text{ right angles} \] From (i), \[ 2 \angle\text{ACB} = 180^\circ \Rightarrow \angle\text{ACB} = 90^\circ \] Hence, the angle in a semicircle is a right angle. Proved.