I saw the following problem on Facebook (figure not drawn to scale):
$\triangle ABC$ is an isosceles triangle with $AB\equiv BC$, $AC\equiv BD$, and $\angle B=20^\circ$. Find $\angle CAD$.
What I did was let $BC\equiv x$ and divide the triangle in two by bisecting $\angle B$. This yielded two right triangles with base $x\cos80^\circ$. Therefore, the base of $\triangle ABC$ has length $2x\cos80^\circ$.
Clearly, $DC=x-2x\cos80^\circ$, and $\angle C=80^\circ$.
Now, to find $AD\equiv y$, I used the law of cosines:
$$ y=\sqrt{(2x\cos80^\circ)^2+(x-2x\cos80^\circ)^2-2(2x\cos80^\circ)(x-2x\cos80^\circ)\cos80^\circ}. $$
Finally, to find $\angle CAD\equiv\theta$, I used the law of sines:
$$\begin{align} \frac{\sin80^\circ}{y}&=\frac{\sin\theta}{x-2x\cos80^\circ}\\ \Longrightarrow\theta&=70^\circ. \end{align}$$
Is my result correct? Is there an easier way to solve this problem?