If $AK$, $BL$ and $CM$ are the altitudes in triangle $ABC$, prove that if $|AM|+|BK|+|CL|=|AL|+|BM|+|CK|$, triangle $ABC$ is isosceles. there is an integersting math contest geometry problem: If $AK$, $BL$ and $CM$ are the altitudes of triangle $ABC$, prove that if $|AM|+|BK|+|CL|=|AL|+|BM|+|CK|$, then triangle $ABC$ is isosceles. It seems like I am stuck here. My idea is to get some polynomial which can be factored into $(a-b)(b-c)(a-c)$, which should equal $0$ of course. In here I haven't made any progress¨yet, I was trying to use Pythagorean theorem, which is not the best choice, since square roots make the computitation a lot harder. Then I used some trigonomteric expressions, but it turned out that there were also pretty useless. 
Could somebody please point me in the right direction? I would be grateful for any hint. Thanks!  
 A: Let $|AM|= x_1, |BK|=x_2, |CL|=x_3$ and $|AL|=y_1, |BM|=y_2, |CK|=y_3$. Then we are given that: $$x_1 + x_2 + x_3=y_1+y_2+y_3$$
Also, as arberavdullahu noted in the comments their products will be equal due to Ceva's theorem, thus:
$$x_1 x_2 x_3 = y_1 y_2 y_3$$
There's one more relation which will see us through. Note that if $H$ is the orthocenter then by Pythagoras theorem:
$$AH^2 + BH^2 + CH^2 = x_1^2 + x_2^2 + x_3^2 + HM^2 + HK^2 +HL^2 = y_1^2 + y_2^2 + y_3^2 + HM^2 + HK^2 +HL^2 $$
Thus, we have:
$$ x_1^2 + x_2^2 + x_3^2 = y_1^2 + y_2^2 + y_3^2$$
And since the sums were given equal, squaring them and using the above relation, we get: $$x_1x_2 + x_2x_3 +x_1x_3= y_1y_2 + y_2y_3 +y_1y_3$$
Note that three quantities $x_1 x_2 x_3$, $x_1x_2 + x_2x_3 +x_1x_3$, $x_1 + x_2 + x_3$ define an unique cubic with roots $x_1, x_2, x_3$. And since all those quantities with $x$ replaced by $y$ is the same, their roots are same as well. Thus we have $\{ x_1, x_2, x_3 \} = \{ y_1, y_2, y_3 \}$. 
Now any one of $x_1 \in \{ y_1, y_2, y_3\}$ will give an isosceles triangle as you can easily check.
