Barycentric is very neat . Quite a few cevians. Length chase. We need to show that $\frac{AK}{CL}=\frac{AE}{EC}$. Now $\frac{AK}{AC} = \frac{\sin \frac{C}{2}}{\sin \frac{A}{2}}$ and $\frac{CL}{AC} = \frac{\sin (90^{\circ} + \frac{C}{2})}{\sin (90^{\circ} - \frac{A}{2})} = \frac{\cos \frac{C}{2}}{\cos \frac{A}{2}}$. Therefore \[\frac{AK}{CL} = \frac{\tan \frac{C}{2}}{\tan \frac{A}{2}} = \frac{\frac{IE}{EC}}{\frac{IE}{EA}} = \frac{EA}{EC}\qquad\square\]

Also quickly dies to bary. Let $D,G$ be the reflections of $A$ over $BI$ and $CI$, respectively and let $F$ be the reflection of $C$ over $BI$. Then $G=(0,b,a-b), F=(a,a-c,0)$ and $D=(0,a-c,c)$. Then $K = AD \cap CI = ( a(a-c), b(a-c), bc)$ and $L = AG \cap CF = (ab, b(c-a), (a-b)(c-a))$. Since $E=(a+b-c,0,-a+b+c)$ we finish by showing that \[\left| \begin{matrix}a+b-c & 0 &-a+b+c \\ a(a-c) & b(a-c) & bc \\ ab & b(c-a) &(a-b)(c-a) \end{matrix} \right|=0.\]

We have $AK \parallel CL$ and a lot of easy angles - so simply just length chase. This is done above.

My solution: Let $(I)$ touch $BC,CA,AB$ at $D,E,F$.$AL$ cuts $CI,BC$ at $M,N$, respectively. It is easy to show that $D,F,M$ are conllinear. We obtain: $\frac{LC}{AK}=\frac{ML}{MA}=\frac{ML}{MN}=\frac{DC}{DN}=\frac{CE}{AE} \Rightarrow K,E,L$ are conllinear. Q.E.D

Attachments:

What a bashy problem! rkm0959 wrote: Let there be an acute triangle $\triangle ABC$ with incenter $I$. $E$ is the foot of the perpendicular from $I$ to $AC$. The line which passes through $A$ and is perpendicular to $BI$ hits line $CI$ at $K$. The line which passes through $A$ and is perpendicular to $CI$ hits the line which passes through $C$ and is perpendicular to $BI$ at $L$. Prove that $E, K, L$ are colinear. Since $AK \parallel CL$, it suffices to show that $\triangle CEL \sim \triangle AEK \iff \frac{AE}{AK} = \frac{CE}{CL}$. Now, we have that $$\frac{AK}{CL} = \frac{AK}{AC} \cdot \frac{AC}{CL} = \frac{\sin{\frac{c}{2}}}{\sin{\frac{a}{2}}} \cdot \frac{\cos{\frac{a}{2}}}{\cos{\frac{c}{2}}} = \frac{\tan{\frac{c}{2}}}{\tan{\frac{a}{2}}} = \frac{EA}{EC}$$