I remember this problem being discussed on the Hyacinthos newsgroup, but I don't succeed to find the proper messages anymore. But why should I if there is a very simple solution? Indeed, from AA' = BB' = CC', it follows that triangle ABC is equilateral. Proof. We first show: Lemma 1. If a > b, then AA' < BB'. Proof. Since the two tangential segments from a point to a circle are equal, we have CA' = CB'. Thus, the triangle A'CB' is isosceles, and we get $\measuredangle A^{\prime}B^{\prime}C = \frac{180^{\circ}-\measuredangle A^{\prime}CB^{\prime}}{2} = \frac{180^{\circ}-C}{2} = 90^{\circ}-\frac{C}{2}$. But < AB'A' = 180 - < A'B'C, and thus $\cos \measuredangle AB^{\prime}A^{\prime} = -\cos \measuredangle A^{\prime}B^{\prime}C = -\cos \left(90^{\circ}-\frac{C}{2}\right) = -\sin\frac{C}{2}$. Also, it is well-known that AB' = s - a and BA' = s - b, where $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Thus, with a > b, we have s - a < s - b, so that AB' < BA'. Now, after the Cosine Law in triangle AA'B', we have $AA^{\prime}\ ^2 = AB^{\prime}\ ^2 + A^{\prime}B^{\prime}\ ^2 - 2\cdot AB^{\prime}\cdot A^{\prime}B^{\prime}\cdot \cos \measuredangle AB^{\prime}A^{\prime} = AB^{\prime} ^2 + A^{\prime}B^{\prime}\ ^2 + 2\cdot AB^{\prime}\cdot A^{\prime}B^{\prime}\cdot \sin\frac{C}{2}$ (in the last step, we used $\cos \measuredangle AB^{\prime}A^{\prime} = -\sin\frac{C}{2}$). Similarly, $BB^{\prime}\ ^2 = BA^{\prime} ^2 + A^{\prime}B^{\prime}\ ^2 + 2\cdot BA^{\prime}\cdot A^{\prime}B^{\prime}\cdot \sin\frac{C}{2}$. Now, comparing these two formulas, we directly see that AA' < BB' because of AB' < BA' (and because of $\sin\frac{C}{2}>0$). Lemma 1 is thus proven. Now, for our problem, if the triangle ABC were not equilateral, we would have one side which greater than the other; WLOG we could assume a > b, and Lemma 1 would yield AA' < BB' in contradiction to our hypothesis AA' = BB' = CC'. So the problem is solved. Darij

Hmm... Maybe some shortl solution? Oh.. I again realized that Darij post is too long We see that $CB'=CA'$, so from cosine theorem \[AA'^2-BB'^2=(AC^2+CA'^2-2AC\cdot CA'\cos C)-(BC^2+CB'^2-2BC\cdot CB'\cos C)=(AC-BC)(AC+BC-2CB'\cos C),\] where $BC,AC>CB'=CA'$, $\cos C\leq 1$. Thus if $AC>BC$ then $AA'^2>BB'^2$.