MathDB
EM_|_AB and similar triangles wanted, <ACB = 90^o

Source: Dutch NMO 2021 p4

December 28, 2021
geometrysimilar trianglesperpendicularright triangle

Problem Statement

In triangle ABCABC we have ACB=90o\angle ACB = 90^o. The point MM is the midpoint of ABAB. The line through MM parallel to BCBC intersects ACAC in DD. The midpoint of line segment CDCD is EE. The lines BDBD and CMCM are perpendicular. (a) Prove that triangles CMECME and ABDABD are similar. (b) Prove that EMEM and ABAB 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("AA", A, SW); dot("BB", B, SE); dot("CC", C, N); dot("DD", D, NW); dot("EE", E, NW); dot("MM", M, S); [/asy]
Be aware: the figure is not drawn to scale.