Problem

Source: Dutch NMO 2021 p4

Tags: geometry, similar triangles, perpendicular, right triangle



In triangle $ABC$ we have $\angle ACB = 90^o$. The point $M$ is the midpoint of $AB$. The line through $M$ parallel to $BC$ intersects $AC$ in $D$. The midpoint of line segment $CD$ is $E$. The lines $BD$ and $CM$ are perpendicular. (a) Prove that triangles $CME$ and $ABD$ are similar. (b) Prove that $EM$ and $AB$ are perpendicular. [asy][asy] unitsize(1 cm); pair A, B, C, D, E, M; A = (0,0); B = (4,0); C = (2.6,2); M = (A + B)/2; D = (A + C)/2; E = (C + D)/2; draw(A--B--C--cycle); draw(C--M--D--B); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, N); dot("$D$", D, NW); dot("$E$", E, NW); dot("$M$", M, S); [/asy][/asy] Be aware: the figure is not drawn to scale.