MathDB

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] 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]

Problem Statement