9.1: Ángulo entre una línea tangente y una cuerda

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Teorema$\PageIndex{1}$

Dejar$\Gamma$ ser un círculo con el centro$O$. Supongamos que la línea$(XQ)$ es tangente a$\Gamma$ at$X$ y$[XY]$ es un acorde de$\Gamma$. Entonces

\[2 \cdot \measuredangle QXY \equiv \measuredangle XOY.\]

Equivalentemente,

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

Prueba

Tenga en cuenta que$\triangle XOY$ es isósceles. Por lo tanto,$\measuredangle YXO = \measuredangle OYX$.

Aplicando el Teorema 7.4.1 a$\triangle XOY$, obtenemos

$2021-02-18 11.12.03.png$

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

Por Lemma 5.6.2,$(OX) \perp (XQ)$, Por lo tanto,

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

Por lo tanto,

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

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