Isosceles triangle $\Delta ABC$ in the figure below: 
$\angle 1+\angle 2=\angle 3+\angle 4,\quad$
$E\in AB,\; D\in AC,\; F=BD\cap CE,$
$BD=CE$.
Prove: $AB=AC$

The exact version figure should look like:

This problem should be a little more difficult than the Steiner-Lehmus Theorem.
 A: We do not loose generality if we place the triangle with the base in $(-1,0),\;(1,0)$.

Let's call the angles $\angle1, \, \cdots, \, \angle 4$ as $\alpha_1,\, \cdots, \, \alpha_4$ (just to better handle them symbolically), then we must have
$$ \bbox[lightyellow] {  
\alpha _{\,1}  + \alpha _{\,2}  = \alpha _{\,3}  + \alpha _{\,4} \quad  \Rightarrow \quad \left\{ \matrix{
  \alpha _{\,4}  - \alpha _{\,1}  = \alpha _{\,2}  - \alpha _{\,3}  = \delta  \hfill \cr 
  \alpha _{\,4}  + \alpha _{\,1}  = \delta  + 2\alpha _{\,1}  = \delta  + \beta  \hfill \cr 
  \alpha _{\,2}  + \alpha _{\,3}  = \delta  + 2\alpha _{\,3}  = \delta  + \gamma  \hfill \cr}  \right.
 } \tag{1}$$
Consider now two segments of length $r$ departing from points $B$ and $C$, and ending at points
$$ \bbox[lightyellow] {  
\eqalign{
  & D = \left( { - 1 + r\cos \left( {\delta  + \beta /2} \right),\;r\sin \left( {\delta  + \beta /2} \right)} \right)  \cr 
  & E = \left( {1 - r\cos \left( {\delta  + \gamma /2} \right),\;r\sin \left( {\delta  + \gamma /2} \right)} \right) \cr} 
 } \tag{2}$$
so that they satisfy the conditions imposed on them for the length, and for the angles (they shall bisect $\beta$ and $\gamma$).
Let's consider then the lines $C,D$ and $B,E$. Their equations are
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  {\rm line}\,{\rm CD}:\;{{x - 1} \over { - 2 + r\cos \left( {\delta  + \beta /2} \right)}} = {y \over {r\sin \left( {\delta  + \beta /2} \right)}} \hfill \cr 
  {\rm line}\,{\rm BE}:{{x + 1} \over {2 - r\cos \left( {\delta  + \gamma /2} \right)}} = {y \over {r\sin \left( {\delta  + \gamma /2} \right)}} \hfill \cr}  \right.
 } \tag{3}$$
and we want their slopes to be:
$$ \bbox[lightyellow] {  
\eqalign{
  & \left\{ \matrix{
  {{r\sin \left( {\delta  + \beta /2} \right)} \over { - 2 + r\cos \left( {\delta  + \beta /2} \right)}} =  - \tan \left( {\delta  + \gamma } \right) \hfill \cr 
  {{r\sin \left( {\delta  + \gamma /2} \right)} \over {2 - r\cos \left( {\delta  + \gamma /2} \right)}} = \tan \left( {\delta  + \beta } \right) \hfill \cr}  \right.\quad  \Rightarrow  \quad (4.a) \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  {{\sin \left( {\delta  + \beta /2} \right)} \over {2/r - \cos \left( {\delta  + \beta /2} \right)}} = \tan \left( {\delta  + \gamma } \right) \hfill \cr 
  {{\sin \left( {\delta  + \gamma /2} \right)} \over {2/r - \cos \left( {\delta  + \gamma /2} \right)}} = \tan \left( {\delta  + \beta } \right) \hfill \cr}  \right.\quad  \Rightarrow  \quad (4.b) \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  \sin \left( {2\delta  + \beta /2 + \gamma } \right) = 2/r\sin \left( {\delta  + \gamma } \right) \hfill \cr 
  \sin \left( {2\delta  + \beta  + \gamma /2} \right) = 2/r\sin \left( {\delta  + \beta } \right) \hfill \cr}  \right. \quad (4.c) \cr} 
 }$$
The system of equations in (4.c) above can be represented as
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  0 \le \beta ,\gamma  < \pi /2 - \delta  \hfill \cr 
  F\left( {\beta ,\;\,\gamma \;;\;\,\delta ,r} \right) = \sin \left( {2\delta  + \beta /2 + \gamma } \right) - 2/r\sin \left( {\delta  + \gamma } \right) \hfill \cr 
  F\left( {\beta ,\;\,\gamma \;;\;\,\delta ,r} \right) = 0 \hfill \cr 
  F\left( {\gamma ,\;\beta \;\,;\,\;\delta ,r} \right) = 0 \hfill \cr}  \right.
 } \tag{5}$$
and since it imposes to be null either the $F(\beta,\,\gamma)$  and its symmetric $F(\gamma,\, \beta)$, 
then, clearly, if there are, the solutions will be $\beta=\gamma$, i.e. the triangle must be isosceles.
Q.E.D.
A: 
I exchanged the postions of $\angle 3, 4$ in the figure above, which should not affect the result. The proof below will use this new figure.
Construct $EG{/\!\!/}BD$ and $DG{/\!\!/BE}$, then $BDGE$ is a parallelogram.
Connect points $G$ and $C$, label angles $\measuredangle 1=\angle EGD=\angle 1$, $\angle 5 =\angle CGD$ and $\angle 6=\angle DGC$.
Then $CE=BD=EG\Rightarrow \measuredangle 1+\angle 6=\angle 3+\angle 5$
Next use the proof by contradition technique:
If $\angle 3=\angle 1$, the result will be easy to prove, so we suppose $\angle 3\ne\angle 1$. Specifically, without the loss of generality, let us suppose:
$\angle 3>\angle 1\Rightarrow \angle 5<\angle 6\tag{1}$
and $\because\angle 1+\angle 2=\angle 3+\angle 4$
On the other hand, the current assumption leads to：
$\angle 3>\angle 1\Rightarrow \angle 4<\angle 2\Rightarrow CD<BE\Rightarrow CD<DG\Rightarrow \angle 6<\angle 5\tag{2}$
(1) conflicts with (2); similarly, the assumption $\angle 3<\angle 1$ will also lead to such contradiction， therefore, we conclude that:
$\angle 1=\angle 3$ ，
therefore $\angle 2=\angle 4$ ，and  $\angle 5=\angle 6$;
and then $\angle 1+\angle 4=\angle 3+\angle 2$ and $AB=AC$
A: I am going to use the original diagram. (The so-called "exact version" diagram has some labels changed, and I think does not help, because it is too easy to assume the answer from it.)
Given $BD=CE$, we are trying to show that the relation governing the angles: $\angle 1 + \angle 2 = \angle 3 + \angle 4$ implies that $AB=AC$. First, observe that the inverse implication is trivially true:
If $AB=AC$, then $BEDC$ is a trapezium (trapezoid in American) with equal base angles and equal diagonals, and therefore $\angle 1 = \angle 3$ and $\angle 2 = \angle 4$ and the equality follows.
But now consider drawing the two lines from the vertex $A$, and fixing the points $C$ and $E$, while allowing the point $D$ to move along the line $AC$, as the point $B$ similarly moves along the line $AE$ to maintain the equality of lengths $EC$ and $BD$. The $\angle 3$ is fixed: consider how the value of $\angle 1 + \angle 2 - \angle 4$ varies.
If $D$ starts close to $A$, then $\angle 1$ will be arbitrarily small, $\angle 2$ will be arbitrarily small or negative, but $\angle 4$ will be significantly large. The value of $\angle 1 + \angle 2 - \angle 4$ will be very small or negative. As $D$ moves downwards, $\angle 1$ and $\angle 2$ are both monotonically increasing, while $\angle 4$ is monotonically decreasing. Therefore, a fortiori, the value of $\angle 1 + \angle 2 - \angle 4$ is monotonically increasing. There must therefore be a unique point at which the angle relation holds. But we have already shown that it holds if $AB=AC$, and therefore the implication goes both ways, and we have proved that given the angle relation, $AB=AC$. QED
