On the sides $AD , BC$ of the square $ABCD$ the points $M, N$ are selected $N$, respectively, such that $AM = BN$. Point $X$ is the foot of the perpendicular from point $D$ on the line $AN$. Prove that the angle $MXC$ is right. (Mirchev Borislav)
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Tags: geometry, right angle, square, equal segments
On the sides $AD , BC$ of the square $ABCD$ the points $M, N$ are selected $N$, respectively, such that $AM = BN$. Point $X$ is the foot of the perpendicular from point $D$ on the line $AN$. Prove that the angle $MXC$ is right. (Mirchev Borislav)