In the first part of the article the proof of the following theorem is given: Let point $S$ be the middle of $AB$ in the triangle $ABC$, point $O$ the intersection of $AB$ and the axis of angle $ACB$, point $P$ the foot of the perpendicular from $C$ on $AB$. If angles $ACS,SCO,OCP,PCB$ are equal, then the angle $BCA$ is the right one. In the second part, the area of right angle triangle using only the length of the axis of the right angle and of the median is derived.