parmenides51 wrote: The midpoint of triangle's side and the base of the altitude to this side are symmetric wrt the touching point of this side with the incircle. Prove that this side equals one third of triangle's perimeter. (A.Blinkov, Y.Blinkov) Consider the triangle $ABC$ and let $H$, $K$, $M$, $K'\in BC$ be the foot of the altitute, tangent point of incircle, middle of $BC$ and the tangent point of excircle. Let also $AK'$ intersects incircle in point $P$. Then $KP$ is the diameter of incercle and $KP=2r$. Hence $AH=3r$. Counting the area of $ABC$, we get $$S=\frac{BC\cdot 3r}{2}=\frac{(AB+BC+AC)}{2}r,$$so $BC=(AB+BC+AC)/3$.