Inscribed Angle Theorem

Theorem

An inscribed angle is equal to half the angle that subtends the same arc. In the figure below, $\angle BAC = \frac{1}{2} \angle BDC$.

As Euclid defined it:

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 III: Proposition $20$)

Euclid's Proof

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 Triangles have 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.

$\blacksquare$

Alternative (incomplete) 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, we can conclude $AD = BD = CD$, hence the triangles $\triangle ADB$ and $\triangle BDC$ are isosceles.

Therefore we may equate angles $\angle DBC = \angle DCB$.

All angles in a triangle add up to 180, so $\angle BDC$ must be a supplement of $\angle DBC + \angle DCB = 2 \angle DCB$.

The angle $\angle ABC$ is right, so 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 $\angle BDC = 2 \angle DAB$, which proves the theorem for this case.

The general case is illustrated below. A diameter is drawn from $A$ through the center $D$ to $E$.

By the previous logic, $\angle BAE = 2 \angle BDE$ and $\angle CAE = 2 \angle CDE$. Subtracting the latter from the former equation obtains the general result.

$\blacksquare$