On the sides $AB$ and $AC$ of the right $\triangle ABC$ ($\angle A=90^{\circ}$) are chosen points $C_{1}$ and $B_{1}$ respectively. Prove that if $M=CC_{1}\cap BB_{1}$ and $AC_{1}=AB_{1}=AM$, then $[AB_{1}MC_{1}]+[AB_{1}C_{1}]=[BMC]$.