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Mathematics LibreTexts

9.1: Angle between a tangent line and a chord

  • Page ID
    23633
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    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 

    截屏2021-02-18 上午11.12.03.png

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

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

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

    Therefore,

    \(2 \cdot \measuredangle QXY \equiv \pi - 2 \cdot YXO \equiv \measuredangle XOY.\)