Let $ABC$ be a triangle with $AB < AC$ and incentre $I$. Let $E$ be the point on the side $AC$ such that $AE = AB$. Let $G$ be the point on the line $EI$ such that $\angle IBG = \angle CBA$ and such that $E$ and $G$ lie on opposite sides of $I$. Prove that the line $AI$, the line perpendicular to $AE$ at $E$, and the bisector of the angle $\angle BGI$ are concurrent.
Problem
Source: Middle European Mathematical Olympiad I-3
Tags: geometry, incenter, symmetry, angle bisector, perpendicular bisector, geometry proposed
pi37
20.09.2014 21:50
By symmetry about $AI$, $AI$ bisects $\angle BIE$. Note that $BA$ bisects $\angle GBI$, so the external bisector of $\angle GBI$ is the perpendicular from $B$ to $BA$, which by symmetry concurs with $AI$ and the perpendicular from $E$ to $EA$. Thus $AI$, the perpendicular from $AE$ to $E$, the perpendicular from $B$ to $AB$, and the bisector of $\angle BGI$ concur at the $G$-excenter of $GBI$.
Number1
21.09.2014 00:13
Say angle bisector at $A$ and perpendicular on $AE$ at $E$ intersect at $X$. Then $\angle XBA = \pi/2$ so $ABXE$ is inscribed in some circle $\mathcal{C}$. Let $IE$ intersect $\mathcal{C}$ at $G'$. Since $X$ halves arc $BE$ which contains $X$ $G'X$ is angle bisector of the angle $\angle BG'I$. Let $\angle IBE = \phi$, $\angle IBA = \alpha$ and $\angle BAI = \gamma$ Since $ \phi+ \alpha+ \gamma = \pi/2$ and $\angle G'AB = \angle G'EB = \angle IEB = \angle EBI = \phi$ and $\angle BG'A = \pi/2 +\gamma$ we have $\angle ABG' = \alpha$. Thus $G'=G$ and we are done.
mathuz
22.09.2014 19:53
We have that $BA$ line is bisector of the triangle $BGI$ and $AI$ is bisector line of the triangle $BEI$. Let the bisector of the angle $BGI$ and $AI$ intersect at a point $P$. Then $P$ is excenter of $BGI$ and $PB \perp AB$. Since $AB=AE$, $ \angle PAB=\angle PAE$ we get that $PE=PB$ and $PE \perp AE$.
Mikasa
22.09.2014 20:48
Note that since $\triangle ABE$ is isosceles and $AI$ is the bisector of $\angle BAE$, $AI$ is the perpendicular bisector of $BE$. Let $P\in AI$ such that $PE\perp AE$. Then $\angle APE=90^{\circ}-\angle PAE=\angle AEB=\angle ABE$ and thus $P\in \odot ABE$. Again, $\angle GBA=\angle IBG-\angle IBA=\angle CBA-\dfrac{1}{2}\angle CBA=\dfrac{1}{2}\angle CBA$ and $\angle IEA=\angle IBA=\dfrac{1}{2}\angle CBA$. So, $\angle GBA=\angle IEA=\angle GEA$. Thus $G\in \odot ABE$ too. Now $\angle BGP=\angle BAP=\angle EAP=\angle EGP$ and thus the line $AI$, the line perpendicular to $AE$ at $E$, and the bisector of the angle $\angle BGI$ are concurrent at $P$, as required.
steppewolf
01.08.2015 22:00
Because $AI$ is the bisector of $\angle BAC$ and $\triangle ABE$ is isosceles we have that $$\angle BIE = 2*(\frac{1}{2}\angle ABC + \frac{1}{2}\angle BAC) = 180^{\circ} - \angle BCA$$So $\angle BIG = \angle BCA$, and because $\angle IBG = \angle CBA$ we have that $\angle BGI = \angle BGE = \angle BAC = \angle BAE$ so $BEAG$ is cyclic. Now if $M$ is the midpoint of arc $BE$ of $\odot BEAG$ not containing $A$, it is well known that $M$ lies on the bisectors of $\angle BGE$ and $\angle BAE$. But $BM$ is the diameter of this circle because $\triangle ABE$ is isosceles and so $AE \perp EM$, so the three lines are concurrent at $M$.
LMat
21.08.2015 19:04
The basic idea behind this problem is that points $A,G,B,E$ are cocylcic. It is then easy to see that all the lines in question go through the midpoint of arc $BE$.
lazizbek42
12.01.2022 09:12
The perpendiculars from $E$ to $AE$ and the perpendiculars from $B$ to $AB$ and $AI$ intersect at one point. let this be the $T$ point.$A,G,B,T,E$ concyclic.