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?


we do it by using the coordinate system like the picture. enter image description here

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.

  • $\begingroup$ This solution seems promising however solve FC=BF is killing me. Thanks anyway. $\endgroup$ – SuperLucky Sep 5 '14 at 14:26
  • $\begingroup$ you re right, it s so complicated that I used a mathmatic software to solve it $\endgroup$ – Martial Sep 5 '14 at 14:39

$\mathbf{Revised\,3}$ (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.


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:

  1. Construct circumcircle $c$ of isosceles $ADE$.
  2. Select point $B$ on the extension of $AD$.
  3. Conduct $BE$. It will intersect circumcircle $c$ at $F$.
  4. Conduct $DF$. It will intersect $AE$ at $C$ (see [*]).
  5. Now apply Desargues (powers of points $B$ and $C$ relative to the circumscribed $c$), using the additional assumptions $AD=AE$ and $BD=FC$.
  6. 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.

  • $\begingroup$ Need some more details on how: (1)+(2)=>BDECisosceles trapezium $\endgroup$ – SuperLucky Sep 5 '14 at 14:11
  • $\begingroup$ @SuperLucky: My apologies. You are right. It needs a bit of explanation. See revised answer. $\endgroup$ – Yiannis Galidakis Sep 5 '14 at 17:00
  • $\begingroup$ Let FG be the perpendicular bisector of the line segment B(G)C such that $\angle BFG = F_1 = F_2 = \angle CFG$. However, extending AF might not cut BC also at G. Not yet anyway. $\endgroup$ – Mick Sep 5 '14 at 17:19
  • $\begingroup$ Hmm. You are both right. I'll see if I can prove it, regardless (that $AF$ is indeed the bisector). $\endgroup$ – Yiannis Galidakis Sep 5 '14 at 17:36

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.

  • $\begingroup$ It is not very clear that why " since line GH is paralell to BC", then "we know that angle GAE = HAD". $\endgroup$ – Mick Sep 7 '14 at 5:17
  • $\begingroup$ @Mick yes that does seem to be kind of the hand wavey part of the proof. I am certain it is true although I cannot exactly put my finger on what theorem says so. I want to think it has something to do with the parallel postulate. $\endgroup$ – KBusc Sep 8 '14 at 12:06
  • $\begingroup$ This is the fun part. Everything seems so obvious but a small piece is missing according to Euclid. $\endgroup$ – Mick Sep 10 '14 at 6:40

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