Inscribed Angle Theorem/Proof 2

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

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