# Inscribed Angle Theorem/Proof 2

## Theorem

An inscribed angle is equal to half the angle that is subtended by that arc.

Thus, in the figure above:

- $\angle ABC = \frac 1 2 \angle ADC$

In the words of Euclid:

*In a circle the angle at the center is double of the angle at the circumference, when the angles have the same circumference as base.*

(*The Elements*: Book $\text{III}$: Proposition $20$)

## Proof

Consider the simplest case that occurs when $AC$ is a diameter of the circle:

Because all lines radiating from $D$ to the circumference are radii and thus equal:

- $AD = BD = CD$

Hence the triangles $\triangle ADB$ and $\triangle BDC$ are isosceles.

Therefore from Isosceles Triangle has Two Equal Angles:

- $\angle DBC = \angle DCB$.

From Sum of Angles of Triangle equals Two Right Angles:

- $\angle BDC$ is a supplement of $\angle DBC + \angle DCB = 2 \angle DCB$.

From Thales' Theorem, $\angle ABC$ is right.

By similar reasoning $\angle DAB$ is the complement of $\angle DCB$.

If $\angle BDC$ is the supplement of twice the complement of $\angle DAB$, then $\angle BDC = 2 \angle DAB$.

That proves the theorem for this case.

$\Box$

The general case is illustrated below.

A diameter is drawn from $A$ through the center $D$ to $E$.

By the previous logic:

- $\angle BDE = 2 \angle BAE$
- $\angle CDE = 2 \angle CAE$

Subtracting the latter from the former equation obtains the general result.

$\blacksquare$