(a)

Beginning with the diagram on the left, Natalia draws $\overline {PB}$ and marks the diagram to show segments that she knows to be congruent because each one is a $\answer [format=string]{radius}$ of the circle.

(b)

Natalia sees that $\triangle APB$ and $\triangle BPC$ are $\answer [format=string]{isosceles}$ triangles, so she marks the figure to show angles that must be congruent.

(c)

In order to do some algebra with these congruent angles, Natalia labels their measures $x$ and $y$, as shown in the picture on the right.

(d)

She writes an equation for the sum of the angles of $\triangle ABC$:

$$\answer {x+(x+y)+y} = 180^\circ$$

(e)

She divides that equation by $\answer {2}$ to conclude that $m\angle ABC = x+y = \answer {90}$ degrees.

In the figure below, $\overline {AC}$ is a diameter of a circle with center $P$. Prove that $z=2x$.

Because $z$ is the measure of an angle exterior to $\triangle \answer {APB}$, it is equal to the sum of the measures of the oppositeadjacentremote interioralternate interior angles. In other words $z=2x$.

Alternatively, without using the exterior angle theorem, one might proceed as follows:

(a)

$\angle APB + x + x = 180^\circ$ because of the angle sum in $\triangle \answer {ABP}$.

(b)

$\angle APB + z = 180^\circ$ because they form a linear pair.

(c)

Then $z = 2x$ by comparing the two equations.

Note: This handles the special case in which the center of the circle lies on one side of the inscribed angle. For the general result, consider two cases: (1) When the center of the circle is in the interior of the inscribed angle; and (2) When the center of the circle is not in the interior of the inscribed angle.