Let $ABC$ be a triangle such that $AB<AC$. The perpendicular bisector of the side $BC$ meets the side $AC$ at the point $D$, and the (interior) bisectrix of the angle $ADB$ meets the circumcircle $ABC$ at the point $E$. Prove that the (interior) bisectrix of the angle $AEB$ and the line through the incentres of the triangles $ADE$ and $BDE$ are perpendicular.
Problem
Source: Romanian TST 2011
Tags: geometry, circumcircle, incenter, perpendicular bisector, geometric transformation, angle bisector
RSM
06.04.2012 22:06
Suppose, $ I $ is the incenter of $ \Delta ADE $ and $ I' $ be the incenter of $ \Delta BDE $. By some simple angle-chasing we can see that, $ \Delta AID $ is similar to $ \Delta DI'E $. So a spiral similarity centered at $ D $ takes $ AI $ to $ EI' $. So $ \Delta DII'%Error. "similar" is a bad command. \Delta DAE $. So $ \angle (II',AE)=\angle IDA=\frac {\angle ACB}{2} $. So $ II' $ is perpendicular to the internal angle-bisector of $ \angle AEB $.
Tafhim
20.10.2017 17:32
Let $I$ and $J$ be the incenters of $\triangle AED$ and $\triangle BED$ respectively.Internal bisector of $\angle AEB$ meets $IJ$ at $F$.Let $M$ and $N$ be the midpoints of arc $ED$ not containing $A$ of $\odot AED$ and arc $ED$ not containing $B$ of $\odot BED$ respectively.Let $\angle BAE=\angle x$. $\angle ADE=\angle BDE=\angle C$; $\angle DAE=\angle A+\angle x$; $\angle AED=180-\angle EAD-\angle ADE$; $\angle BEA=180-\angle C$; $\angle BED=\angle BEA-\angle AED=180-\angle C-\angle B+\angle x=\angle A+\angle x$ $\implies$ $\triangle AED$ and $\triangle EBD$ are similar, and $BE$ and $AE$ are tangent to $\odot AED$ and $\odot BED$ respectivelly. $MN$ is the perpendicular bisector of $ED$. $\angle IED=\frac{\angle AED}{2}=\frac{\angle EBD}{2}=\angle NBD=\angle NED$ $\implies$ E,I,N are collinear. Similarly E,J,M are collinear. Now,$\angle IMJ=\angle AME=\angle ADE=\angle BDE=\angle BNE=\angle JNI$ $\implies$ $I,J,M,N$ are cyclic. So,$\angle EIF=\angle EIJ=\angle EMN=\frac{\angle EMD}{2}=\frac{180-\angle EAD}{2}=\frac{180-\angle A-\angle x}{2}=90-\frac{\angle A}{2}-\frac{\angle x}{2}$. $\angle FEI=\angle FEA-\frac{\angle AED}{2}=\frac{\angle BEA}{2}-\frac{\angle AED}{2}=\frac{180-\angle C}{2}-\frac{\angle B-\angle x}{2}=90-\frac{\angle B}{2}-\frac{\angle C}{2}+\frac{\angle x}{2}$. So,$\angle FEI+\angle EIF=90$ $\implies$ $\angle EFI=90$.