In triangle ABC we have ∠ACB=90o. 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]
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]Be aware: the figure is not drawn to scale. geometrysimilar trianglesperpendicularright triangle