Inscribed Angle Theorem
Thus, in the figure above:
- $\angle ABC = \frac 1 2 \angle ADC$
In the words of Euclid:
- 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.
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.
Consider the simplest case that occurs when $AC$ is a diameter of the circle:
- $AD = BD = CD$
Therefore from Isosceles Triangle has Two Equal Angles:
- $\angle DBC = \angle DCB$.
- $\angle BDC$ is a supplement of $\angle DBC + \angle DCB = 2 \angle DCB$.
By similar reasoning $\angle DAB$ is the complement of $\angle DCB$.
That 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:
- $\angle BDE = 2 \angle BAE$
- $\angle CDE = 2 \angle CAE$
Subtracting the latter from the former equation obtains the general result.