NL-NWO 2018 Problem 4

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 $|AG| =|BF|$ .

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);