The incircle of a triangle $ABC$ has centre $I$ and touches sides $AB, BC, CA$ in points $C_1, A_1, B_1$ respectively. Denote by $L$ the foot of a bissector of angle $B$, and by $K$ the point of intersecting of lines $B_1I$ and $A_1C_1$. Prove that $KL\parallel BB_1$. proposed by L. Emelyanov, S. Berlov
Problem
Source: 239MO 2004, grade 8-9, problem 2
Tags: geometry, Russia, bisector
grobber
12.12.2004 00:28
Let's start with the easy ones . Let $T=BL\cap A_1C_1$. $TB_1LK$ is cyclic, so $\frac{IK}{IL}=\frac{IT}{IB_1}\ (*)$. What we want to show is $\frac{IK}{IL}=\frac{IB_1}{IB}$. Combining this with $(*)$, we see that what we need is $IB_1^2=IT\cdot IB$. However, this is clear from the right triangle $IC_1B$. We have $IB_1^2=IC_1^2=IT\cdot IB$.
Megus
12.12.2004 14:38
Excuse my lack of geometry knowledge but when a quadriteral is cyclic and what it gives [I don't understand $(*)$] ?
grobber
12.12.2004 14:40
It's when the four points lie on the same circle. $(*)$ takes place because of the equality of angles of the triangles.
Megus
12.12.2004 14:45
Thanks - drawing proper picture also helps.
jayme
05.04.2014 15:32
Dear Mathlinkers, and without any calculation? what do you think? Sincerely Jean-Louis
ThirdTimeLucky
06.04.2014 22:08
An outline: (1) $ BK $ meets $ AC $ at its midpoint, $ M $. (2) $ MI $ meets $ BB_1 $ at its midpoint, $ X $. (3) The pencil $ (KX,KB_1,KL,KM) $ is harmonic. So, $ KL \parallel BB_1 $.
jayme
07.04.2014 08:42
Dear, yes, this is also my proof... Sincerely Jean-Louis