Prove $\triangle BAC$ is a right-angled isosceles triangle Suppose in the figure below, $AD\  \text{//} \ BC$, $BD=BC$, $CD=CE$, 
 and $ABCD$ is a trapezoid; $\measuredangle ABD=15°$
Prove that: $\triangle BAC$ is a right-angled isosceles triangle. 


Let $F$ the intersection of lines $BA$ and $CD$, $\angle BCD=  x$ and $BC=L$.
Then 
$$\angle BFC=  x-\frac{\pi}{12}. \tag{1}$$
Applying some trig we get:
$$CD=2L\cos x \tag{2}$$
$$DE =4L(\cos x)^2 \tag{3} $$.
Applying sine rule in triangle $DAE$ we get:
$$DA=\frac{4L(\cos x)^2\sin x}{-\sin(3x)} \tag{4} $$
The similarity of triangles $FDA$ and $FCB$ produces:
$$FD=\frac{2L\cdot DA \cos x}{L -DA}  \tag{5}$$
Applying sine rule in triangle $FDB$ we get:
$$FD=\frac{L \sin(\frac{ \pi}{12})}{ \sin( x-\frac{ \pi}{12})} \tag{6}$$
Using $(4)$ and making $(5)$ and $(6)$ equal, we get:
$$\sin(\frac{\pi}{12})(7-8(\sin x)^2)=-8(\cos x)^3\sin( x-\frac{ \pi}{12}) \tag{7}$$
After some trig relations, we get:
$$\tan(x-\frac{ \pi}{12})=\tan x (8(\sin x)^2-7)$$
Solving for $x$ we get:
$$x = \frac{5\pi}{12}$$
Thereafter it is easy to conclude that angle $BAC$ is a right angle and $\triangle ABC$ is isosceles.
