Inscribed Angle Theorem/Proof 1

Proof

 * Euclid-III-20.png

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.