請教一題

以 \(\overline{AB}\) 為直徑做一個半圓,圓心為 \(O\), \(C\)為半圓上的一點,若 \(\overline{OC}^2=\overline{AC}\times \overline{BC}\), 則下列關於 \(\angle CAB\) 的度數何者正確?
(A)        必為 \(30^\circ \)
(B)        必為 \(75^\circ \)
(C)        可能為 \(30^\circ \)或\(60^\circ \)
(D)        可能為 \(15^\circ \)或\(75^\circ \)

請教一題,如附件,謝謝!!

\(\begin{align}
  & \Delta ABC=\frac{1}{2}\times \overline{AC}\times \overline{BC} \\
& \Delta ABC=\Delta AOC+\Delta BOC \\
& =\frac{1}{2}\times \overline{OC}\times \overline{OA}\times \sin \angle AOC+\frac{1}{2}\times \overline{OC}\times \overline{OB}\times \sin \angle BOC \\
& =\frac{1}{2}\times \overline{OC}\times \overline{OC}\times \sin \angle AOC+\frac{1}{2}\times \overline{OC}\times \overline{OC}\times \sin \angle AOC \\
& ={{\overline{OC}}^{2}}\times \sin \angle AOC \\
&  \\
& \sin \angle AOC=\frac{1}{2} \\
& \angle AOC={{30}^{{}^\circ }}\ or\ {{150}^{{}^\circ }} \\
& \angle CAB={{15}^{{}^\circ }}\ or\ {{75}^{{}^\circ }} \\
\end{align}\)

請問如果是國中生沒學過三角函數,此題要如何解?謝謝!!

跟上面差不多
作\(\overline{CD}\)垂直\(\overline{AB}\)於D，注意：D可能在\(\overline{OA}\)或\(\overline{OB}\)上
\(\begin{align}
  & \Delta ABC=\frac{1}{2}\times \overline{AC}\times \overline{BC} \\
& \Delta ABC=\Delta AOC+\Delta BOC \\
& =\frac{1}{2}\times \overline{OA}\times \overline{CD}+\frac{1}{2}\times \overline{OB}\times \overline{CD} \\
& =\frac{1}{2}\times \overline{OC}\times \overline{CD}+\frac{1}{2}\times \overline{OC}\times \overline{CD} \\
& =\overline{OC}\times \overline{CD} \\
&  \\
& \overline{CD}=\frac{1}{2}\overline{OC} \\
& \angle COD={{30}^{{}^\circ }} \\
& \angle CAB={{15}^{{}^\circ }}\ or\ {{75}^{{}^\circ }} \\
\end{align}\)