There is a triangle $ABC$ where $|CB|=a$, $|AC|=b$ and medians of these sides intersect at a right angle. Find |AB|.

I don't know how to use a right angle in this problem. I have a idea to link a middle of $|CB|$ with $|AC|$, let $K,L$ be a centre of these sides and we have $2|KL|=|AB|$


We've the known medians theorem in geometry: the three medians to the sides of a triangle meet at one point which divides each median in a ratio of $\;1:2\;$ , with the longest segment always on the vertex side.

Thus, calling $\;M\;$ to the intersection point of the two medians, we can put


with $\;D=\;$ the midpoint of $\;BC\;$ , and $\;E=\;$ the midpoint of AC.

Using now Pythagoras on the triangle $\;\Delta AMB\;$ we get:


But on $\;\Delta BMD\;$ we get:


and on $\;\Delta AME\;$ we get


so solving the quadratic system of two unknowns above, we get:



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