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. 

 A: 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.
A: According to bigant146's request, I'll try to make a solution as synthetic as possible.
Denote $\angle DBC=\angle ECD=\alpha$.
Mark a point $P$ on $BC$ so that $BP=BE$.
Choose a point $Q$ so that $BEQP$ is a rhombus.
Clearly, $Q$ lies on the symmetry axis of the triangle $BDC$.
We have $\angle DEQ=\angle DBC=\angle BDA$ and
$$
  \frac{DE}{EQ}=\frac{DE}{EB}=\frac{DA}{BC}=\frac{DA}{DB},
$$
so the triangles $DEQ$ and $ADB$ are similar; thus $\angle DQE=15^\circ$.
Due to the symmetry, $\angle CQP=\angle DQE=15^\circ$ as well.
Thus in an isosceles triangle $DQC$ ($DQ=QC$) we have $\angle DQC=\alpha+30^\circ$.
Draw now a circle $\omega$ centered at $C$ with radius $CD=CE$.
If $\alpha<30^\circ$, then $\angle DQE>\angle DCE/2$,
so $Q$ lies inside $\omega$;
on the other hand, $\angle DQC<60^\circ$, so $QC>DC$,
and hence $Q$ lies outside $\omega$ --- a contradiction.
A similar contradiction is obtained when $\alpha>30^\circ$; thus $\alpha=30^\circ$.

The rest is straightforward: we have $\angle ABC=\angle ABD+\angle DBC=15^\circ+\alpha=45^\circ$
and $\angle ACB=\angle DCB-\angle DCA=(90^\circ-\alpha/2)-\alpha=45^\circ$.
A: It is straightforward to check that the data of the question, including the condition $\angle ABD=\frac1{12}\pi$, are consistent with the value $\frac14\pi$ for $\angle ABC$ and $\angle ACB$. However, this fact alone does not rule out other possibilities for $\angle ABC$ (and other angles in the figure). To establish the unique result, we consider the more general problem where $\theta:=\angle ABD$ is variable rather than fixed as $\frac1{12}\pi$.
Choose the unit of length so that $|BC|=|BD|=1$, and write $x:=|CD|$. Since the isosceles triangles $DBC$ and $DCE$ share the base angle $\angle BDC$, they are similar, and so $|ED|=x^2$. Then $|EB|=1-x^2$ and, by the similarity of triangles $EBC$ and $EDA$, we have $$|AD|=\frac{x^2}{1-x^2}.$$Now write $\phi:=\angle DBC$. Then $\angle BDA=\phi$ and $\angle BAD=\pi-(\theta+\phi)$. Applying the sine rule to triangle $BAD$ gives $|BD|/\sin(\theta+\phi)=|AD|/\sin\theta$, or $$\left(\frac1{x^2}-1\right)\sin\theta=\sin\theta\cos\phi+\cos\theta\sin\phi.$$Dividing through by $\cos\theta$, and using the fact that $x=2\sin\frac12\theta$, so that $x^2=2(1-\cos\phi)$, some rearrangement gives $$\tan\theta=f(\phi):=\frac{2(1-\cos\phi)\sin\phi}{2\cos^2\phi-1}.$$As $\phi$ increases from $0$ towards $\frac14\pi$, each variable factor in the numerator of the above expression for $f(\phi)$ increases (from $0$), while the denominator decreases (from $1$ down to $0$). Hence $f$ is an increasing function in this range. We do not need to consider $\phi$ boyond $\frac14\pi$, because then $\tan\theta<0$ and $\theta>\frac12\pi$, and we are ultimately only interested in the case $\theta=\frac1{12}\pi$. Thus, in the range of interest for $\phi$, our function $f$ is increasing, and so $\tan\theta$ and hence $\theta$ is increasing with respect to phi. Conversely, it follows that $\phi$, and hence $\theta+\phi$, is increasing with respect to $\theta$ in $(0\;\pmb,\,\frac12\pi)$. Consequently there is a unique value of $\theta+\phi$ corresponding to each $\theta$ in this range—in particular for $\theta=\frac1{12}\pi$.
Using the formula $$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta},$$with $\tan2\theta=\tan\frac16\pi=1/\surd3$, and solving the resulting quadratic equation for $\tan\theta$, we get $\tan\frac1{12}\pi=2-\surd3$. This value is also obtained by calculating $f(\frac16\pi)$. Hence $\phi=\frac16\pi$ when $\theta=\frac1{12}\pi$, and then $\angle ABC=\frac1{12}\pi+\frac16\pi=\frac14\pi$. Setting $\gamma:=\angle ACB$, noting that $\angle BCD=\gamma+\phi=\angle BDC$, and summing the angles in triangle $BCD$ gives $3\phi+2\gamma=\pi$, or $\gamma=\frac14\pi$. Thus $\angle ACB=\frac14\pi$ too, and finally $\angle BAC=\frac12\pi$.
