Problem

Source: Dutch NMO 2018 p4

Tags: geometry, equal segments



In triangle $ABC, \angle A$ is smaller than $\angle C$. Point $D$ lies on the (extended) line $BC$ (with $B$ between $C$ and $D$) such that $|BD| = |AB|$. Point $E$ lies on the bisector of $\angle ABC$ such that $\angle BAE = \angle ACB$. Line segment $BE$ intersects line segment $AC$ in point $F$. Point $G$ lies on line segment $AD$ such that $EG$ and $BC$ are parallel. Prove that $|AG| =|BF|$. [asy][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("$A$", A, SW); dot("$B$", B, NE); dot("$C$", C, NE); dot("$D$", D, NE); dot("$E$", E, N); dot("$F$", F, N); dot("$G$", G, SW); [/asy][/asy]