equal segments wanted, old school geometry style

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equal segments wanted, old school geometry style

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Source: Dutch NMO 2018 p4

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#1 h Sep 7, 2019, 10:33 AM • 1 Y

Y by Adventure10

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$", G, SW); [/asy]$

This post has been edited 2 times. Last edited by nsato, Feb 13, 2023, 9:07 PM

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#2 h Sep 7, 2019, 12:31 PM • 2 Y

Y by Adventure10, Mango247

$AG \parallel BF$ ( $\angle ABF = \angle B/2 = \angle DAB$ ).
$AEBG$ - isoscales trapezoid ( $\angle AGE = \angle EBC = \angle EBA \implies AEBG$ - cyclic).
$\angle AEF = \angle AFE$ ( $\triangle BCF \sim \triangle BAE$ )
$\implies AFBG$ - parallelogram.

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#4 h Dec 5, 2019, 8:21 AM • 1 Y

Y by Adventure10

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 $\implies AG=FB$ .

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#5 h Mar 30, 2020, 11:17 AM

Y by

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 $EGA$ $\implies AG=BF$ $Q.E.D$

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