# 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. ### Thébault's Problems

• Thébault's Problem I
• Thébault's Problem II
• Thébault's Problem III
• Y. Sawayama's Lemma
• Jack D'Aurizio Proof of Sawayama's Lemma
• Y. Sawayama's Theorem
• Thébault's Problem III, Proof (J.-L. Ayme)
• Circles Tangent to Circumcircle
• Thébault's Problem IV
• A Property of Right Trapezoids
• Excircles Variant of Thébault's Problem III
• In the Spirit of Thebault I
• Dao's Variant of Thebault's First Problem
• 