بدون کاستن از کلیت مسئله فرض کنیم:
$BC = 2 , A=C=\theta$
پس:
$\quad AD = \tan \theta \quad BD = CD = 1$
$\begin{equation}
\frac{\sin \theta}{BE} = \frac{\sin\left(180^\circ -\frac{ 3\theta}{2}\right)}{BC}
\end{equation}$
$\begin{align*}
\frac{\sin \theta}{2 \tan \theta} & = \frac{\sin\left(180^\circ- \frac{3 \theta}{2}\right)}{2} \\
\cos \theta & = \sin \left(\frac{3\theta}{2}\right) \\
\cos \frac{\theta}{2} & = \sin \frac{3\theta}{2} \quad \Rightarrow \quad 1 -\sin^2 \frac{ \theta}{2} = 3 \sin \frac{\theta}{2} - 4\sin^3\frac{ \theta}{2}
\end{align*}$
$\begin{align*}
4x^3 - 2x^2 - 3x + 1 = 0 \\
(4x^2 + 2x - 1)(x - 1) = 0 \quad \Rightarrow \quad
\begin{cases}
x_1 = 1 \\
x_2 = \frac{\sqrt{5} - 1}{4}
\end{cases}
\end{align*}$
$x_1 = 1 \quad \Rightarrow \quad \sin \frac{\theta}{2} = 1 \quad \Rightarrow \quad \frac{\theta}{2} = 90^\circ \quad \Rightarrow \quad \theta = 180^\circ \quad (\text{غقق})$
$x_2 = \frac{\sqrt{5} - 1}{2} \quad \Rightarrow \quad \sin \frac{\theta}{2} = \frac{\sqrt{5} - 1}{2} \quad \Rightarrow \quad \frac{\theta}{2} = 18^\circ \quad \Rightarrow \quad \theta = 36^\circ$
$$\hat{C} = \theta = 36^\circ$$
