Let $\omega$ be the circumcircle of triangle $ABC$. Tangent lines to $\omega$ at points $A$ and $C$ intersect at $K$. Line $BK$ intersects $\omega$ for the second time at $M$. On the line $BC$ point $N$ is chosen such that $\angle BAN = 90$. Line $MN$ intersects $\omega$ for the second time at $D$. Prove that $BD=BC$ P. Chernikova