9.1: Angle between a tangent line and a chord

Theorem $\PageIndex{1}$

Let $\Gamma$ be a circle with the center $O$. Assume the line $(XQ)$ is tangent to $\Gamma$ at $X$ and $[XY]$ is a chord of $\Gamma$. Then

$$2 \cdot \measuredangle QXY \equiv \measuredangle XOY.$$

Equivalently,

$\measuredangle QXY \equiv \dfrac{1}{2} \cdot \measuredangle XOY$ or $\measuredangle QXY \equiv \dfrac{1}{2} \cdot \measuredangle XOY + \pi.$

Proof

Note that $\triangle XOY$ is isosceles. Therefore, $\measuredangle YXO = \measuredangle OYX$.

Applying Theorem 7.4.1 to $\triangle XOY$, we get

$\begin{array} {rcl} {\pi} & \equiv & {\measuredangle YXO + \measuredangle OYX + \measuredangle XOY \equiv} \\ {} & \equiv & {2 \cdot \measuredangle YXO + \measuredangle XOY.} \end{array}$

By Lemma 5.6.2, $(OX) \perp (XQ)$, Therefore,

$\measuredangle QXY + \measuredangle YXO \equiv \pm \dfrac{\pi}{2}.$

Therefore,

$2 \cdot \measuredangle QXY \equiv \pi - 2 \cdot YXO \equiv \measuredangle XOY.$