引用:
三角形ABC
以BC為直徑做圓
AB交圓於D
AC交圓於E
假設三角形AED面積為1
四邊形DECB面積為t
求cos角BAC
\(\begin{align}
& AD\times AB=AE\times AC \\
& \frac{AD}{AC}=\frac{AE}{AB} \\
& \frac{\Delta ADE}{\Delta ABC}=\frac{AD\times AE}{AB\times AC}=\frac{1}{t+1} \\
& {{\left( \frac{AD}{AC} \right)}^{2}}=\frac{1}{t+1} \\
& \cos \angle BAC=\frac{AD}{AC}=\sqrt{\frac{1}{t+1}} \\
\end{align}\)