MathDB

4

Part of 2018 Dutch Mathematical Olympiad

Problems(1)

equal segments wanted, old school geometry style

Source: Dutch NMO 2018 p4

9/7/2019
In triangle ABC,∠AABC, \angle A is smaller than ∠C\angle C. Point DD lies on the (extended) line BCBC (with BB between CC and DD) such that ∣BD∣=∣AB∣|BD| = |AB|. Point EE lies on the bisector of ∠ABC\angle ABC such that ∠BAE=∠ACB\angle BAE = \angle ACB. Line segment BEBE intersects line segment ACAC in point FF. Point GG lies on line segment ADAD such that EGEG and BCBC are parallel. Prove that ∣AG∣=∣BF∣|AG| =|BF|.
[asy] unitsize (1.5 cm);
real angleindegrees(pair A, pair B, pair C) { real a, b, c; a = abs(B - C); b = abs(C - A); c = abs(A - B); return(aCos((a^2 + c^2 - b^2)/(2*a*c))); };
pair A, B, C, D, E, F, G;
B = (0,0); A = 2*dir(190); D = 2*dir(310); C = 1.5*dir(310 - 180); E = extension(B, incenter(A,B,C), A, rotate(angleindegrees(A,C,B),A)*(B)); F = extension(B,E,A,C); G = extension(E, E + D - B, A, D);
filldraw(anglemark(A,C,B,8),gray(0.8)); filldraw(anglemark(B,A,E,8),gray(0.8)); draw(C--A--B); draw(E--A--D); draw(interp(C,D,-0.1)--interp(C,D,1.1)); draw(interp(E,B,-0.2)--interp(E,B,1.2)); draw(E--G);
dot("AA", A, SW); dot("BB", B, NE); dot("CC", C, NE); dot("DD", D, NE); dot("EE", E, N); dot("FF", F, N); dot("GG", G, SW); [/asy]
geometryequal segments