Hard Geometry Question Problem:
In triangle $ABC$, $D$ and $E$ are two points on side $BC$ such that $BD = CE$ and $\angle BAD = \angle CAE$. Prove that triangle $ABC$ is isosceles.
I am having a lot of trouble with this question; what I have done so far is figure out that I need to use a translation that takes $B$ to $E$ and $D$ to $C$. Solutions would be greatly appreciated.


Let $b:=\overrightarrow{AB}$, $c:=\overrightarrow{AC}$ and $A$ be the origin. Let $\frac{BD}{BC}=\frac{CE}{BC}=x$. Then
$$\overrightarrow{AD}=xc+(1-x)b,\quad
\overrightarrow{AE}=xb+(1-x)c,\\
\cos\angle BAD=\frac{
(\overrightarrow{AB}\cdot \overrightarrow{AD})}{
|\overrightarrow{AB}|\cdot |\overrightarrow{AD}|},\quad
\cos\angle CAE=\frac{
(\overrightarrow{AC}\cdot \overrightarrow{AE})}{
|\overrightarrow{AC}|\cdot |\overrightarrow{AE}|},
$$
thus (equating the cosine squares)
$$\frac{(b(xc+(1-x)b))^2}{b^2(xc+(1-x)b)^2}=
\frac{(c(xb+(1-x)c))^2}{c^2(xb+(1-x)c)^2}$$
$$(b(xc+(1-x)b))^2c^2(xb+(1-x)c)^2=(c(xb+(1-x)c))^2b^2(xc+(1-x)b)^2$$
$$c^2(xbc+(1-x)b^2)^2(x^2b^2+2x(1-x)bc+(1-x)^2c^2)=\\
b^2(xbc+(1-x)c^2)^2(x^2c^2+2x(1-x)bc+(1-x)^2b^2)$$
$$c^2(x^2(bc)^2+2x(1-x)(bc)b^2+(1-x)^2b^4)(x^2b^2+2x(1-x)bc+(1-x)^2c^2)=\\
b^2(x^2(bc)^2+2x(1-x)(bc)c^2+(1-x)^2c^4)(x^2c^2+2x(1-x)bc+(1-x)^2b^2)$$
However, I'm more convinient with feeding this thing to WolframAlpha at this point (scalars $A:=bc,\,B:=b^2,\,C:=c^2$), see here:
$$(x - 1) x^2 (-(b^2 - c^2)) (c^2 b^2 - (bc)^2) (-2 (bc) x + c^2 x - c^2 + b^2 x - b^2)=0$$
So it's either

*

*$x=1$ but then $D=C,\,E=B$ or

*$x=0$ but then $D=B,\,E=C$ or

*$AB^2=AC^2$ or

*$\cos^2\angle CAB=
\frac{
((\overrightarrow{AB}\cdot \overrightarrow{AC}))^2}{
|\overrightarrow{AB}|^2\cdot |\overrightarrow{AC}|^2}=
\frac{(bc)^2}{b^2c^2}=1$ so $\cos\angle CAB=\pm 1$ and the triangle $\triangle ABC$ is degenerate or

*$x(b-c)^2=b^2+c^2$ but then the cosines ($\cos\angle CAE$, $\cos\angle DAB$) signs are different so the angles can't be equal. This thing did come up because we equated the squared cosines, not the cosines themselves.

So out of all the possibilities it's left only $AB^2=AC^2$ i.e. $|AB|=|AC|$, QED.
