Inscribed Angle Theorem/Proof 1
Theorem
An inscribed angle is equal to half the angle that is subtended by that arc.
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.
(The Elements: Book $\text{III}$: Proposition $20$)
Proof
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.
$\blacksquare$
Historical Note
This proof is Proposition $20$ of Book $\text{III}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions