Fermat -Euler wrote: Let $ABC$ be a triangle, and let the angle bisectors of its angles $CAB$ and $ABC$ meet the sides $BC$ and $CA$ at the points $D$ and $F$, respectively. The lines $AD$ and $BF$ meet the line through the point $C$ parallel to $AB$ at the points $E$ and $G$ respectively, and we have $FG = DE$. Prove that $CA = CB$. Let a=BC, b=CA, c=AB. We have BF/AD=(BF/FG)/(AD/DE)=(c/a)/(c/b)=b/a, but it's easy to see from the formulae of the type la=2bc*cos(A/2)/(b+c) that the order of the bisectors of a triangle is inverse to that of the sides (if a>=b>=c then la<=lb<=lc). If b>a then BF/AD=b/a>1, which is false, so b<=a => contradiction. We do the same for b<a, so the only possible situation is a=b, Q.E.D.

Let $ I$ be the incenter, and let $ CI$ intersect $ AB$ at $ P$. Since $ GE \parallel AB$, we have $ \triangle ABD \sim \triangle DEC$ and $ \triangle ABF \sim \triangle FCG$. Thus, $ \frac {BD}{DC} = \frac {AD}{DE}$ and $ \frac {AF}{FC} = \frac {BF}{FG}$. Since $ DE = FG$, dividing the two we have $ \frac {BD}{DC} \cdot \frac {CF}{FA} = \frac {AD}{BF}$. But by Ceva, $ \frac {BD}{DC} \cdot \frac {CF}{FA} = \frac {PB}{PA} = \frac {BC}{CA}$ (angle bisector theorem). So $ \frac {BC}{CA} = \frac {AD}{BF}$. But if $ BC \neq CA$, then this would mean that lengths of the angle bisectors are in the same order as the lengths of the sides they intersect, a contradiction by the generalization of the Steiner-Lehmus Theorem in the link http://www.cems.uvm.edu/~cooke/sl.pdf. Thus, $ BC = CA$.

Never seen such an easy geometry in the ISLs atleast. Wlog let $a \ge b$.Note that $CF=\frac{ab}{a+c},\angle{GCA}=A,\angle{FGC}=\frac{B}{2}$ So application of sine rule in $\triangle{CFG}$ yeids $\frac{GF}{sinA}=\frac{ab}{(a+c)sin\frac{B}{2}}$ Similarly sine rue in $\triangle{CDE}$ yeilds $\frac{ED}{sinB}=\frac{ab}{b+c}$ Dividing the two relations we get $\frac{(a+c)b}{(b+c)a}=\frac{sin\frac{A}{2}}{sin\frac{B}{2}} \ge 1$ by our assumption. This gives $\frac{(a+c)b}{(b+c)a} \ge 1$ $\Rightarrow ab+bc \ge ab+ac \Rightarrow b \ge a$ So we obtain $b=a$

OK, maybe some logical order. Everyone up, who made assumption that $a\geq b$ and didn't check the case $a<b$ has only a half of solution. By the way you can't have an assumption that $a=b$, so you can't make the case $a\geq b$, because you have to prove, that $a=b$. How can the proof be correct, when you prove something supposing it. Correct solution is when you show, that both inequalities $a>b$ and $a<b$ are contradiction, so there must be $a=b$.

WolfusA wrote: OK, maybe some logical order. Everyone up, who made assumption that $a\geq b$ and didn't check the case $a<b$ has only a half of solution. By the way you can't have an assumption that $a=b$, so you can't make the case $a\geq b$, because you have to prove, that $a=b$. How can the proof be correct, when you prove something supposing it. Correct solution is when you show, that both inequalities $a>b$ and $a<b$ are contradiction, so there must be $a=b$. Your first complaint is incorrect because you can make the assumption $a\ge b$ because the two cases are exactly symmetric, hence the "Wlog" (just flip your diagram). In addition, your second complaint is incorrect because supposing $a\ge b$ is not the same thing as $a=b$. Obviously at least one of $a\ge b, a\le b$ is true, and since they are symmetric you may assume one of them. You should learn some logic yourself before criticizing others.

I don't mean that WLOG part is incorrect. So if you like so this equality in assumption check yourself that this $p \vee q \implies p$ is tautology. And put here $p: a=b$ and $q: a>b$. Anyway the second case (after WLOG) is $a<b$ (not $b\geq a)$.

WolfusA wrote: I don't mean that WLOG part is incorrect. So if you like so this equality in assumption check yourself that this $p \vee q \implies p$ is tautology. And put here $p: a=b$ and $q: a>b$. Anyway the second case (after WLOG) is $a<b$ (not $b\geq a)$. I'm not sure exactly what you're saying (I'm pretty sure the first thing you said isn't a sentence), but $p \vee q \implies p$ is not a tautology. $a\ge b$ does not always imply $a=b$, that is absurd. Also, the point of WLOG is that there are no cases, we only have to check one of the cases by symmetry. The reason I said $b\ge a$ rather than $b>a$ is to preserve the symmetry, so that the WLOG is valid.

Sorry, but I made a mistake with this "tautology". I withdraw from accusations.

If $AC \neq BC$, WLOG, assume that $\angle CAB > \angle CBA$, then we have \[BC > AC, BG > AE.\]Since $GF=DE$, we have $FB=BG-GF >AE-GF=AE- DE=DA$, so \[GF=\frac{FB \cdot GC}{AB}=\frac{FB \cdot BC}{AB} > \frac{DA \cdot AC}{AB}=\frac{DA \cdot CE}{AB}=DE\]which contradicts $GF=DE$, thus, $AC=BC$. $\blacksquare$