Prove that $\sin 2\alpha=2 \sin \alpha \cos\alpha$ In this triangle

$AD=AC=1$, $BC=a$, $BAC=2\alpha$
I thought $\sin 2\alpha=a$, but I don't know how to continue.
 A: drop a perpendicular from $A$ to $DC$ and call the foot of the perpendicular $E.$ then $$DE =  EC = \cos \alpha, DC = 2 \cos \alpha.$$   now look at the right triangle $CBD$ with hypotenuse $2\cos \alpha$ and the $\angle BDC = \alpha,$  therefore the opposite side $BC$ is $$ BC = 2\cos \alpha \sin \alpha.$$  you can also look at the right angle triangle $ABC$ with hypotenuse $AC = 1, \angle BAC = 2\alpha$ which give the opposite side $$BC = \sin 2\alpha $$
therefore $$ BC = \sin 2\alpha = 2\sin \alpha \cos \alpha $$
A: Try using an identity. $\sin 2\alpha = \sin \alpha + \alpha = \sin \alpha \cos \alpha + \sin\alpha\cos\alpha = 2\sin\alpha\cos\alpha $ 
A: $e^{2\alpha i}=\cos{2\alpha}+i\sin{2\alpha}=(e^{\alpha i})^2=(\cos{\alpha}+i \sin{\alpha})^2=\cos^2{\alpha}-\sin^2{\alpha}+2i \sin{\alpha}\cos{\alpha}$ Now equating the imaginary part gives the formula.
A: Consider a right triangle $ABC$ where the angle $A$ is right and the angle $B$ is $\alpha$. Let $M$ be the middle point of $BC$.
Then, the angle $AMC$ is $2\alpha$. Now apply on the triangle $AMC$ the law of sines:
$$\frac{\sin2\alpha}{AC}=\frac{\sin(90-\alpha)}{\frac12BC}$$
Since $AC=BC\sin\alpha$ and $\sin(90-\alpha)=\cos\alpha$ the identity follows.
Remark: This proof is only valid for acute angles.
