Inscribed Angle Theorem

An inscribed angle is equal to half the angle that subtends the same arc. In the figure below, $$\angle BAC = \frac{1}{2} \angle BDC$$.
 * [[Image:IncsribedAngle.PNG|300px]]

Proof
Consider the simplest case that occurs when $$AC$$ is a diameter of the circle:


 * [[Image:InscribedAngle15.PNG|300px]]

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.


 * [[Image:IncsribedAngle2.PNG|300px]]