PROBLEM
Let $\triangle ABC$ be isosceles with $AB=AC$. On the extension of side $BC$, a point $D$ is considered such that $C$ belongs to side $BD$ and $BD=AD$. If the bisector $\angle ACD$ forms with the side $AB$ an $\angle AEC$ with a measure of $30$, calculate the measures of the angles of the triangle $AED$.
WHAT I THOUGHT OF
First of all, the drawing:
Ok, so as you can see I noted $\angle BAC=x$ and $\angle CAD=y$
Using the fact that $AB=AC$ and $BD=AD$ we can simply show that $30=\frac{y}{2}$ $=>$ $y=60$, $x=20$
We found out that $\angle EAD=80$.
Now, I don't know how to calculate the other angles.
I thought of using the intern bisector or extern bisector theorem or Ceva's theorem, but it didn't send me any useful ideas. Hope one of you can help me!
Best Answer
Proceeding from the point you stopped by, let's assume $\angle MED=\alpha.$ Then, $\angle MDE=70^{\circ}-\alpha$.
By the law of sines in $\triangle AED$, we have:
$$\frac{\sin \alpha }{\sin 30^{\circ}}=\frac{DM}{AM}\times \frac{AE}{ED}.$$
But, since the segment $CX$ is the bisector, we get that: $\frac{DM}{AM}=\frac{DC}{AC}.$
Hence,
$$\frac{\sin \alpha }{\sin 30^{\circ}}=\frac{DC}{AC} \times \frac{AE}{ED} \\= \frac{\sin 60^{\circ}}{\sin 20^{\circ}}\times \frac{\sin (70^{\circ}-\alpha)}{\sin 80^{\circ}}= \frac{\sin 60^{\circ}}{\sin 20^{\circ}}\times \frac{\cos (20^{\circ}+\alpha)}{\sin 80^{\circ}}.$$
On the other hand, $\sin 20^{\circ}\sin 40^{\circ}\sin 80^{\circ}=\frac{\sqrt 3}{8}$ (look at this ), so:
$$\frac{\sin \alpha}{\cos (20^{\circ}+\alpha)}=\frac{\sin 40^{\circ}}{\cos 60^{\circ}}=\frac{\sin 40^{\circ}}{\cos (20+40)^{\circ}} \implies \alpha=40^{\circ}.$$