Exercise 40 Page 729

For the first part, let's consider a circle and an inscribed angle $\angle FGH$ such that $\overgroup{FH}$ is a semicircle.

By the Inscribed Angle Theorem, we get that $m\angle G=\frac{1}{2}m\overgroup{FJH}.$ Because $\overgroup{FJH}$ is a semicircle, its measure is $180^{\circ }.$ The inscribed angle is a right angle. Conversely, let's consider a circle and an inscribed angle $\angle FGH$ such that $m\angle FGH=90^{\circ }.$ Applying once more the Inscribed Angle Theorem, we can find the measure of $\overgroup{FJH}.$ From the above we conclude that $\overgroup{FJH}$ is a semicircle and $\overline{FH}$ is a diameter.

Paragraph Proof

Theorem $10.8$

An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle.

Proof: Let $\angle FGH$ be an inscribed angle such that $\overgroup{FH}$ is a semicircle. By the Inscribed Angle Theorem, we get that $m\angle FGH=\frac{1}{2}m\overgroup{FJH}.$ Since $\overgroup{FJH}$ is a semicircle, its measure is $180^{\circ }.$ Therefore, $m\angle FGH=\frac{1}{2}(180^{\circ })$ or $90^{\circ }$ which means that $\angle FGH$ is a right angle.
Conversely, let $\angle FGH$ be an inscribed angle such that $m\angle FGH=90^{\circ }.$ The Inscribed Angle Theorem gives us $m\angle FGH=\frac{1}{2}m\overgroup{FJH},$ which implies that $m\overgroup{FJH}=180^{\circ }.$ In consequence, $\overgroup{FJH}$ is a semicircle and $\overline{FH}$ is a diameter.