A non-isosceles triangle $\triangle ABC$ has its incircle tangent to $BC, CA, AB$ at points $D, E, F$. Let the incenter be $I$. Say $AD$ hits the incircle again at $G$, at let the tangent to the incircle at $G$ hit $AC$ at $H$. Let $IH \cap AD = K$, and let the foot of the perpendicular from $I$ to $AD$ be $L$. Prove that $IE \cdot IK= IC \cdot IL$.
Problem
Source: 2016 KMO Senior #2
Tags: geometry, incenter
Hypernova
12.11.2016 12:03
Simple problem using similarity and cyclic.
lminsl
12.11.2016 13:06
This is also Junior KMO 2016 P2.
SAM8
12.11.2016 14:48
Let $GG\cap BC\ \cap EF={Q}$. Then $Q,L,I$ are collinear.Because $IL=\dfrac{ID^2}{IQ} $ it's enough to prove that $\triangle IEK \sim \triangle IQC$ .But $\angle {KIE}=180-\angle {HIE}=180-\angle {GDE}=90-\dfrac{\angle{C}}{2} +90 -\dfrac{\angle{GQD}}{2}=\angle {CIQ}$ and $\angle {IKE}=90-\angle{DGE}=90-\angle{EFD}=\dfrac{\angle{C}}{2}=\angle{ICQ}$
khyeon
13.11.2016 08:01
Fitting angle very useful
CeuAzul
21.06.2017 16:02
Fast & easy proof: $\triangle GIK \cong \triangle EIK \Longrightarrow \angle IEK=\angle IGK=\angle IDK \Longrightarrow \text{D,K,I,E concyclic }$ With $\text{E,I,D,C concyclic } \Longrightarrow \text{E,I,K,D,C concyclic} \Longrightarrow \angle IKL=\angle ICD $ Since $IL \perp LK,IE \perp EC \Longrightarrow \triangle ILK \sim \triangle IEC \Longrightarrow IE \cdot IK= IC \cdot IL $
jayme
30.06.2017 12:10
Dear Mathlinkers, for the cocyclicity, we can use this nice result http://jl.ayme.pagesperso-orange.fr/Docs/Simplicity%201.pdf p. 9-10 which avoid some angle chasing... Sincerely Jean-Louis
Mogmog8
22.11.2021 08:02
Notice $DKIE$ is cyclic as $$\measuredangle IDG=\measuredangle DGI=\measuredangle IEK.$$Thus, $$\measuredangle IKL=\measuredangle EKI=\measuredangle ECI$$so $\triangle IKL\sim\triangle ICE.$ $\square$