The Problems of Ukrainian Authors

Let $AB=c$, $BC=a$, $FM=ML=c$ ($M$ is a middle of $AC$), $AF=LC=x$. According to known formula find length of median (Fig.25)



Fig.25

$$\begin{array}{rl}B{M}^2=m^2& =\frac{1}{4}(2c^2+2a^2-(2b+2x{)}^2)=\frac{1}{4}(2(2b+x{)}^2+2x^2-(2b+2x{)}^2)=\\ & =\frac{1}{4}(8b^2+8bx+4x^2-4b^2-8bx-4x^2)=b^2.\end{array}$$

Therefore in the $△FBL$ median is half of base, in which it has marked, that's why this triangle is right and $\angle FBL=90^{\circ }$.