Inscribed Angle Theorem

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


 * 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.

Alternative (incomplete) proof
Consider the simplest case that occurs when $AC$ is a diameter of the circle:


 * InscribedAngle15.PNG

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.


 * IncsribedAngle2.PNG