Tangents to Circle from Point subtend Equal Angles at Center/Corollary

Theorem

Let $\CC$ be a circle.

Let $P$ be a point in the exterior of $\CC$.

Let $PA$ and $PB$ be tangents to $\CC$ from $P$ touching $\CC$ at $A$ and $B$ respectively.

Then $OP$ is a bisector of $\angle APB$.

Proof

From Tangents to Circle from Point subtend Equal Angles at Center:

$\angle OPA = \angle OPB$

from which follows the result.

$\blacksquare$

Sources

• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): circle $(4) \ \text {(c)}$
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): circle $(4) \ \text {(c)}$