Find BC/AE

Let $BC=1$ . We note that $\triangle ABC$ is isosceles, therefore $AB=AC=\dfrac {\csc 20^\circ}2$ . By sine rule , $\dfrac {AD}{AB} = \dfrac {\sin \angle DBA}{\sin \angle ADB} \implies AD = \dfrac {\sin 10^\circ}{\sin 150^\circ} \times \dfrac {\csc 20^\circ}2 = \dfrac {\sec 10^\circ}2$ . Let $\angle ADC = \theta$ . By sine rule again:

$\begin{aligned} \frac {AC}{\sin \theta} & = \frac {AD}{\sin (120^\circ - \theta)} \\ \frac {\frac 12 \csc 20^\circ}{\sin \theta} & = \frac {\frac 12 \sec 10^\circ}{\sin (60^\circ + \theta)} \\ \frac {\sin(60^\circ + \theta)}{\sin \theta} & = \frac {\sin 20^\circ}{\cos 10^\circ} =\frac {\sin 160^\circ}{\sin 80^\circ} = \frac {\sin (60^\circ +100^\circ)}{\sin 100^\circ} \\ \implies \theta & = 100^\circ \end{aligned}$

By sine rule again, $AE = \dfrac {\sin \theta}{\sin(160^\circ - \theta)}AD = \dfrac {\sin 100^\circ}{\sin 60^\circ} \times \dfrac 1{2 \cos 10^\circ} = \dfrac {\sin 100^\circ}{\frac {\sqrt 3}2} \times \dfrac 1{2 \sin 100^\circ} = \dfrac 1{\sqrt 3}$

Therefore, $\dfrac {BC}{AE} =\sqrt 3 \approx \boxed{1.732}$ .

Let $|\overline {BC}|=a, |\overline {AB}|=|\overline {AC}|=b, |\overline {AE}|=x, \angle {BCD}=α$ . Then $a=2b\cos 70\degree, |\overline {AD}|=2b\sin 10\degree=b\dfrac{\sin (70\degree-α)}{\sin (50\degree+α)}\implies \tan α=\dfrac{\sin 70\degree-2\sin 10\degree\sin 50\degree}{\cos 70\degree+2\sin 10\degree\cos 50\degree}\implies α=50\degree\implies x=\dfrac{b\sin 20\degree}{\sin 60\degree}\implies \dfrac{a}{x}=\dfrac{2b\sin 20\degree\sin 60\degree}{b\sin 20\degree}=\sqrt 3=\boxed {1.732}$ .