Prove two triangles are congruent 
I have found a problem form internet and got stucked trying to proof or disproof it.
It says:
Given $AD=AE$, $BF=FC$, prove $\triangle ABE\cong\triangle ACD$
Update 1
The @Matrial's solution seems very promising however solving $FC=BF$ is killing me, wonder if there were solutions, say, more Euclidean?
 A: we do it by using the coordinate system like the picture.

Suppose: $A=(0,0),D=(a,0),B=(b,0)$, because $AD=AE$, so $E=(a\cos(\theta),a\sin(\theta))$. $E$ is a point on the line $AC$, so we can propose a coefficient $\lambda$ that satisfies $AC=\lambda\cdot AE$ and $0<\lambda<1$, then $C=(\lambda a\cos(\theta),\lambda a\sin(\theta))$.
Now we can get the function of the two line $CD$ and $BE$, the result is
$$CD:y=\frac{\lambda\sin(\theta)}{\lambda \cos(\theta)-1}(x-a)$$
$$BE: y=\frac{a\sin(\theta)}{a\cos(\theta)-b}(x-b)$$
Now we calculate the intersection of these two lines and get
$$F:x_F=\frac{a[\lambda\cos(\theta)(a-b)-(\lambda-1)b]}{a-\lambda b}\\y_F=\frac{a\lambda\sin(\theta)(a-b)}{a-\lambda b}$$
Now we use the condition $FC=BF$, and have the equation about the coefficient $\lambda$
$$(b-x_F)^2+y_F^2=(\lambda a\cos(\theta)-x_F)^2+(\lambda a\sin(\theta)-y_F)^2$$
Solve it and eliminate the solution no real and the solution $\lambda>1$, we have the unique solution
$$\lambda = \frac{b}{a}$$
Now we can say that $AC=AB$ and your equality is proven.
A: $\mathbf{Revised\,3}$ (to reflect the OP's concern)

And all the above/previous $\mathbf{erased}$ with much gusto, after recognising that any geometric proof of it will likely be circular, so it will lead one to endless circles... (probably because of the symmetry of the particular congruence).
I think, but I am not sure, that it can be solved another way. If we can prove that the quadrilateral $DFEA$ is inscribable, then one may try using the corresponding ratios one gets from the power of the points $B$ and $C$ relative to the circumscribed circle of that quadrilateral. Perhaps the assumptions are enough to make those powers equal, hence all the goodies therein.

$\mathbf{Addendum\,\,2}$
And indeed, it looks like Desargues' with the additional assumptions $AD=AE$ and $BF=FC$, which should be immediate.
Sketch of proof for the general case:


*

*Construct circumcircle $c$ of isosceles $ADE$.

*Select point $B$ on the extension of $AD$.

*Conduct $BE$. It will intersect circumcircle $c$ at $F$.

*Conduct $DF$. It will intersect $AE$ at $C$ (see [*]).

*Now apply Desargues (powers of points $B$ and $C$ relative to the circumscribed $c$), using the additional assumptions $AD=AE$ and $BD=FC$.

*Under the two assumptions, 5) implies $F$ lies on the bisector of $A$, and everything else follows.


[*] The degenerate case of the above construction, is when $DE$ is a diameter of $c$. In this case the inscribed quadrilateral will be a square, with both its diagonals diameters, but in this case assumptions $BF=FC$ don't make sense, so this case is rejected. If you omit this case, point $C$ exists.
Needs a lot of work to see fully. If you don't see it, I will try to add more to this.
A: Construct BC then Construct a line parallel to BC through point A.
Next extend BE until at meets the parallel line from the previous step and do the same with line CD, we'll call these points where they meet the parallel line G and H respectively
Now we have angle FBC = FCB = AGE = AHD (by ulternate interior angles of parallel lines cut by a transversal.)
Now since line GH is paralell to BC we know that angle GAE = HAD 
Now by AAS we have trianlge AGE congruent to AHD
Now since angle ADH = AEG then we know that angle ADC = AEB( supplementary angles to congruent angles)
And now we have triangle AEB congruent to ADC by ASA
Sorry about the bad formatting but I'm in quite a rush. 
