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Let $AD$, $BE$ and $CF$ be the angle bisectors of triangle $ABC$. Let the points $D$, $E$ and $F$ lie on $BC$, $AC$ and $AB$, respectively. Let $I$ denote the incenter of $ABC$.

Prove that if the sum of the areas of the inside triangles $ICE$, $IAF$ and $IBF$ equals half the area of triangle $ABC$, then $ABC$ is isosceles. In other words:

$$\text{area}(ICE) + \text{area}(IAF) + \text{area}(IBF) = \frac{1}{2} \text{area} (ABC)\Rightarrow ABC~ \text{is isosceles}.$$

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  • $\begingroup$ If I'm not mistaken, this is actually true if we only take $\overline{BE}$ to be a bisector, and we let $I$ be any (interior) point of $\overline{BE}$ except the midpoint. Points $D$ and $F$ are irrelevant, because the area criterion simplifies to $$|\triangle ICE|+|\triangle IAB| = |\triangle IAE|+|\triangle ICB|$$ From here, I'll leave as a hint to invoke the Angle Bisector Theorem to establish proportions among the sub-triangle areas. $\endgroup$ – Blue May 14 '18 at 7:29
  • $\begingroup$ @Blue Suppose that $S_{ABE} < S_{ABC}/2$. If $I$ is close to $E$, $S_{IAB} + S_{ICE} < S_{ABC}/2$; if $I$ is close to $B$, $S_{IAB} + S_{ICE} > S_{ABC}/2$. Therefore, we can find a point where the area criterion holds in a non-isosceles triangle if we don't have further constraints on $I$. $\endgroup$ – Maxim May 14 '18 at 11:32
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    $\begingroup$ @Blue: Thanks indeed - got your hint and could solve the problem directly! $\endgroup$ – Holbi May 14 '18 at 13:39
  • $\begingroup$ @Maxim: The problematic $I$ is the midpoint of $\overline{BE}$, which is why I wrote that my revision of the result holds except in that case. :) $\endgroup$ – Blue May 14 '18 at 16:19
  • $\begingroup$ @Blue You're right, you had that case covered. So apparently this can be generalized even further: in an arbitrary triangle, for an arbitrary $E$ on $AC$, the area condition holds if either $AE = CE$ (then $I$ is arbitrary) or $IB = IE$ (then $S_{IAB} + S_{ICE} - S_{ABC}/2$ is a linear function of the position of $I$ on $BE$ and therefore has exactly one zero). $\endgroup$ – Maxim May 14 '18 at 17:16
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Using $S = a b \sin(\gamma)/2$ and the angle bisector theorem, we have $$\frac {S_1} {S_2} = \frac {S_{IAB}} {S_{IBC}} = \frac {AB} {BC} = k, \\ \frac {S_4} {S_3} = \frac {S_{IAE}} {S_{ICE}} = \frac {AE} {CE} = k, \\ (S_1 + S_3) - (S_2 + S_4) = (k S_2 + S_3) - (S_2 + k S_3) = \\ {(k - 1) (S_2 - S_3)} = 0.$$ The first possibility is $k = 1$, giving the required result (then the difference $S_2 - S_3$ is arbitrary). The second possibility is $$S_2 = S_3 \Rightarrow BC = CE, \\ S_1 = S_4 \Rightarrow AB = AE, \\ AB + BC = AE + CE = AC,$$ giving a degenerate triangle.

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  • $\begingroup$ Thank you for elaborating on this aspect, much appreciated. $\endgroup$ – Holbi May 14 '18 at 13:40
  • $\begingroup$ You can get there slightly-more-easily by writing $$S_1+S_3=S_2+S_4 \quad\to\quad kS_2+S_3=S_2+kS_3\quad\to\quad (k-1)(S_2-S_3)=0$$ Thus, either $k=1$ (and we're done), or $S_2=S_3$. In the latter case, $I$ must be the midpoint of $\overline{BE}$; requiring $\overline{IA}$ and $\overline{IC}$ to be angle bisectors is one particular way of denying that possibility, via the degeneracy argument you give. $\endgroup$ – Blue May 14 '18 at 16:05
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    $\begingroup$ @Blue Thanks, I got rid of the fractions, they were unnecessary. The second part is I think unchanged, one still needs to prove why $S_2 = S_3$ is impossible. $\endgroup$ – Maxim May 14 '18 at 16:26

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