9.1: Angle between a tangent line and a chord
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
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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,
\(2 \cdot \measuredangle QXY \equiv \pi - 2 \cdot YXO \equiv \measuredangle XOY.\)