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]
Problem
Source: Dutch NMO 2018 p4
Tags: geometry, equal segments
07.09.2019 15:31
$AG \parallel BF$ ($\angle ABF = \angle B/2 = \angle DAB$). $AEBG$ - isoscales trapezoid ($\angle AGE = \angle EBC = \angle EBA \implies AEBG$ - cyclic). $\angle AEF = \angle AFE$ ($\triangle BCF \sim \triangle BAE$) $\implies AFBG$ - parallelogram.
05.12.2019 11:21
In $\Delta CBA$, we have $BF$ as the internal angle bisector of $\angle CBA$. This implies that $\frac{CB}{BA}=\frac{CF}{FA}$. Now given that $AB=BD$ $\implies \frac{CB}{BD}=\frac{CF}{FA}$. This in turn implies that in $\Delta ACD$, we have $FB||AD$. Now let $\angle CBF=\angle FBA=y$. Also let that $EG$ and $AB$ intersect at $H$. Given that $FB||AG$ and $CB||EG$, we have $\angle FBA=\angle BAG=y$ and $\angle CBA=\angle BHG=2y$. This in turn implies that $\angle EGA=y$. Therefore $\angle EBA=\angle EGA=y$, which implies that $EBGA$ is a cyclic quadrilateral. Now $\angle EAB=\angle ACB=z$. Now since $EBGA$ is a cyclic quadrilateral, we have $\angle EAB=\angle EGB=z$. Again $\angle EGB=\angle GBD=z\implies \angle GBD=\angle ACD=z\implies GB||AC$. Therefore, in quadrilateral $AGBF$ we have, $AG||FB$ and $BG||FA$. Therefore $AGBF$ is a parallelogram $\implies AG=FB$.
30.03.2020 14:17
Here is nice solution using angle chasing. $\angle ABE=\angle CBF$ (because we have bisector), $\angle BFC= \angle BEA=\angle AFE=\angle FEA$ , so triangle $AFE$ is isosceles triangle with $AF=AE$. We will now use parallel lines. Since $FB||AG$, we have that $\angle EGA =\angle CDA= \frac{\angle ABC}{2}=\angle ABF=\angle FBC$, because triangle $ABD$ is isosceles. $\angle GEA=180^\circ -\angle EGA-\angle EAB-\angle BAG=180^\circ-\angle ABF-\angle FCB-\angle ABF=180^\circ-\angle ABC-\angle FCB=\angle BAC=\angle BAF$, so triangle $ABF$ is congruent with triangle $EGA$ $\implies AG=BF$ $Q.E.D$