Let ABC be a triangle. r and s are the angle bisectors of ∠ABC and ∠BCA, respectively. The points E in r and D in s are such that AD‖ and AE \| CD. The lines BD and CE cut each other at F. I is the incenter of ABC. Show that if A,F,I are collinear, then AB=AC.
Problem
Source: 38th Brazilian MO (2016) - First Day, Problem 1
Tags: geometry, incenter, Parallel Lines, Brazilian Math Olympiad 2016
23.11.2016 20:04
Since AEID is parallelogram AI bisects DE and since A, I , F are collinear we have that FI bisects DE. In \triangle FDE: BE,DC,FN are concurrent at I (N is midpoint of DE) thus by Cevas theorem BC || DE (since DN = EN). Now we have that AI also bisects BC and since AI is angle bisector of \angle BAC we have AB = AC.
28.02.2017 20:46
Since AE || CD and AD || BE , so, \frac {FC}{CE} = \frac {FI}{IA} = \frac {FB}{BD} . Thus, BC || DE. Simple angle chasing proves that ADBCE is cyclic and hence the result follows.
09.04.2017 00:13
18.08.2017 03:43
We use barycentric coordinates. As D lies in \angle ACB angle bisector, we have that D=(a:b:t) for some t, therefore AD is given by the equation y:z=b:t. We know that BI is given by x:z=a:c. Since BI\|AD, we have that BI\cap AD, is the point at infinity, hence the solutions of the system y:z=b:t ; x:z=a:c must be with x+y+z=0, then, by easy calculations, we find that t=\frac{-cb}{a+c} \Rightarrow D=(a:b:\frac{-cb}{a+c}) and similarly E=(a:\frac{-cb}{a+b}:c). Taking BD \cap CE, we find that F=(a(a+b)(a+c):-cb(a+c):-cb(a+b)). We know that F\in AI \Leftrightarrow (-cb)(a+c):(-cb)(a+b)=b:c\Leftrightarrow ac+c^{2}=ab+b^{2}\Leftrightarrow a(c-b)=-(b+c)(c-b)\Leftrightarrow b=c (Since if b\neq c \Rightarrow 0<a=-(b+c)<0, contradiction), as desired.
23.10.2018 16:17
Clearly AI bisects DE, and DC,BE,FI are concurrent,so BC is parallel to DE, so BC also is bisected by AI, but AI is angle bisector of angle BAC,so we are done.
25.10.2018 20:35
Just simple angle chasing: \angle DAB = \angle ABI = \angle CAE = \angle ACD. And we know BI and CI are angle bisectors \Rightarrow \angle ABC = \angle ACB \Rightarrow AB=AC. Hence Proved. P.S: I wonder why the unnecessary observation DE || BC is required, as seen in all the other solutions.
29.05.2020 15:03
mather007 wrote: Just simple angle chasing: \angle DAB = \angle ABI = \angle CAE = \angle ACD. And we know BI and CI are angle bisectors \Rightarrow \angle ABC = \angle ACB \Rightarrow AB=AC. Hence Proved. P.S: I wonder why the unnecessary observation DE || BC is required, as seen in all the other solutions. You assumed that E&D lies on the circumcircle which you should prove. In fact you used the fact that AB=AC to prove it!!
29.07.2020 08:11
From the parallel conditions we have that ADIE is a paralelogram. Since F lies in the perpendicular bisector of DE, FD=FE. Now from Thales Theorem, \frac{AD}{IB}=\frac{AF}{IF} and \frac{AE}{CI}=\frac{AF}{IF} so \frac{AD}{IB}=\frac{AE}{IC} and from angle-chasing, \angle DAE=\angle DIE=\angle BICso \triangle BIC~\triangle DAE. So a homothety from F sents \triangle BIC to the isosceles triangle \triangle DAE. So we get that \triangle BIC is an isosceles triangle therefore so is \triangle BAC