Inscribed Angle Theorem

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 1

Let $ABC$ be a circle, let $\angle BEC$ be an angle at its center, and let $\angle BAC$ be an angle at the circumference.

Let these angles have the same arc $BC$ at their base.

Let $AE$ be joined and drawn through to $F$.

Since $EA = EB$, then from Isosceles Triangle has Two Equal Angles we have that $\angle EBA = \angle EAB$.

So $\angle EBA + \angle EAB = 2 \angle EAB$.

But from Sum of Angles of Triangle equals Two Right Angles we have that $\angle BEF = \angle EBA + \angle EAB$.

That is, $\angle BEF = 2 \angle EAB$.

For the same reason, $\angle FEC = 2 \angle EAC$.

So adding them together, we see that $\angle BEC = 2 \angle BAC$.

Now consider the point $D$, from which we have another angle $\angle BDC$.

Let $DE$ be joined and produced to $G$.

Similarly, we prove that $\angle GEC = 2 \angle EDC$.

Then $\angle GEB = 2 \angle EDB$.

Therefore $\angle BEC$ which remains is equal to $2 \angle BDC$.

Hence the result.

$\blacksquare$

Proof 2

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.

$\blacksquare$