Problem

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Tags: geometry, circumcircle, trapezoid, geometry proposed



$ABC$ is a right triangle with $\angle C=90$,$CD$ is perpendicular to $AB$,and $D$ is the foot,$\omega$ is the circumcircle of triangle $BCD$,$\omega_{1}$ is a circle inside triangle $ACD$,tangent to $AD$ and $AC$ at $M$ and $N$ respectively,and $\omega_{1}$ is also tangent to $\omega$.prove that: (1)$BD*CN+BC*DM=CD*BM$ (2)$BM=BC$