Tangents to Circle from Point subtend Equal Angles at Center
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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:
- $\angle OPA = \angle OPB$
Corollary
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
Let $O$ be the center of $\CC$.
Construct $OA$ and $OB$.
From Radius at Right Angle to Tangent:
- $PA \perp OA$ and $PB \perp OB$
and so $\angle OAP = \angle OBP$ which equals a right angle.
Consider the right triangles $\triangle OAP$ and $\triangle OBP$.
We have:
- $OA = OB$ by definition of radius of circle
- $\angle OAP = \angle OBP$
- $OP$ is their common hypotenuse.
Hence by Triangle Right-Angle-Hypotenuse-Side Congruence:
- $\triangle OAP \cong \triangle OBP$
and so $OPA = \angle OPB$.
$\blacksquare$
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): circle $(4) \ \text {(b)}$
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): circle $(4) \ \text {(b)}$