# A Lemma on the Road to Sawayama

Below we give an additional proof of Y. Sawayama's Lemma.

Given triangle \(ABC\), construct its circumcircle \(C(O)\) with center \(O\). Now take any point \(D\) on side \(BC\) and draw line \(AD\). Construct circle \(C(O_{1})\), with center \(O_1\), internally tangent to \(C(O)\), \(BC\), and \(AD\). Let \(I\) be the incenter of ABC, \(C(O_1)\) be tangent to \(BC\) at \(L\) , and \(AD\) at \(J\) . Then \(I\), \(L\), and \(J\) are collinear.

### Proof

The proof is due to Oleg Golberg. I thank Sohail Farhangi for bringing the proposition and the proof to my attention.

Let \(LI\) intersect \(C(O_{1})\) at \(J'\). We will show that \(J'\) and \(J\) coincide. Let \(KL\) intersect \(C(O)\) for the second time at \(M\). We can see from homothety with center \(K\) that \(M\) is the midpoint of arc \(BC\) in \(C(O)\). Also from homothety we can see that chords \(KL\) and \(KM\) subtend arcs of equal angles inscribed in \(C(O_{1})\) and \(C(O)\), respectively. Note also that \(I\) lies on \(AM\). So now

\(\angle LJ'K = \angle MAK = \angle IAK = 180^{\circ}-\angle KJ'I,\)

which tells us that \(A, I, J', K\) are concyclic. Observe next that \(\angle MCB = \angle MBC = \angle MKC\), implying that triangles \(MLC\) and \(MCK\) are similar, so that \(MC^2 = ML\cdot MK\). Noting the well known fact that \(MI = MC\), we have \(MI^2 = ML\cdot MK\) from which triangles \(MIL\) and \(MKI\) are similar, and, therefore, \(\angle MIK =\angle MLI\) and \(\angle KLI = \angle KIA = \angle KJ'A\), which tells us that \(AJ'\) is tangent to \(C(O_{1})\) as desired.

|Contact| |Front page| |Content| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

63043950 |