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.

Suppose $$AB \ne AC$$

Construct $$M$$ such that $$AM$$ as the median of $$\triangle ABC$$
Locate $$B',D$$' such that $$\angle ABM = \angle AMB'$$

$$\triangle AMB' \cong \triangle AMB$$ and $$\triangle AB'D' \cong \triangle ABD$$

$$B'D' \cong CE$$ implies $$AE \parallel AC$$

But that is not the case as the two lines clearly intersect at A.

Hint:

Suppose $$\triangle ABC$$ is not isosceles but $$\angle BAD = \angle CAE$$. Can you show that $$BD \not = CE$$?

Try placing a flipped copy of $$\triangle ABC$$ on top of the original such that the new $$AB'$$ lies on the old $$AC$$ and the new $$AC'$$ lies on the old $$AB$$

For triangle $$ABC$$, draw the circumscribed circle.

From $$\angle BAD=\angle CAE$$, we get that the corresponding arc lengths $$BD$$ and $$CE$$ are equal, hence $$BD=CE$$.

Then the arc lengths $$CE'D'$$ and $$BD'E'$$ are also equal, hence $$\angle D'BC=\angle E'CB$$.

It follows that $$\triangle{D'BD}\cong\triangle{E'CE}$$, hence $$\angle BDD'=\angle CEE'$$.

But then $$\angle BDA=\angle CEA'$$, so triangles $$ABD$$ and $$ACE$$ are similar, hence in triangle $$ABC$$, we have $$\angle B=\angle C$$.

Therefore triangle $$ABC$$ is isosceles.

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

1. $$x=1$$ but then $$D=C,\,E=B$$ or
2. $$x=0$$ but then $$D=B,\,E=C$$ or
3. $$AB^2=AC^2$$ or
4. $$\cos^2\angle CAB= \frac{ ((\overrightarrow{AB}\cdot \overrightarrow{AC}))^2}{ |\overrightarrow{AB}|^2\cdot |\overrightarrow{AC}|^2}= \frac{(bc)^2}{b^2c^2}=1$$ so $$\cos\angle CAB=\pm 1$$ and the triangle $$\triangle ABC$$ is degenerate or
5. $$x(b-c)^2=b^2+c^2$$ 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.