Sawayama’s Lemma and Thebault’s circles

This paper compiles a collection of geometric proofs and related constructions that center around Sawayama’s Lemma and Thebault’s theorem. The document begins with the presentation and proof of the Shooting Lemma, which establishes a relationship between a chord in a circle, a tangent circle, and the midpoint of the larger arc. Using this foundational lemma, the proofs of Sawayama’s Lemma and Verrier’s Lemma follow, demonstrating the collinearity of specific points associated with inscribed triangles and tangent circles. The final section extends these results to prove Thebault’s theorem, generalizing the principles of Sawayama’s and Verrier’s Lemmas to a broader context involving two tangent circles.

1 The Shooting Lemma

Proof. The proof is quite trivial, simply consider the homothety centered at $E$ , which transforms $ω$ into $Ω$ . Then, $B$ is mapped to $B ′$ and $C$ to $C ′$ , where $B ′$ and $C ′$ are the intersections of $CE$ and $BE$ with the tangent from $M$ to $Ω$ respectevely.

Then, $∠BCM = ∠CM C ′$ due to $CB ∥ C′B′$ , and $∠CBM = ∠CM C′$ because $C ′B ′$ touches $Ω$ and finally $∠CBM = ∠BM B′$ . Thus, $∠CM C ′ = ∠BM B′$ , however due to $Ω$ touching $C ′B ′$ we know that $∠CEM = ∠CM C′$ and $∠BEM = ∠BM B ′$ , consequently $∠CEM = ∠M EB$ . In other words $M E$ is the bisector of $∠CEB$ which means that $M$ is the middle of the larger arc $⌢ BC$ . ■

In fact, due to $∠BEM = ∠DBM$ we conclude that $(EDB )$ touches $BM$ , consequently,

And due to $M$ being the middle of $B⌢C$ it must be that $CM = M B$ , thus, $M B2 = M C ⋅M B$ . Combining these results we get a nice formula,

The figure can also show a lot of interesting and fundemental properties if one performs an inversion centered at the point $M$ with a radius of $M B = M C$ . Then through this process $Ω ↔ BC$ and because $ω$ must continue to touch $inv(BC )$ and $inv (Ω )$ (in other words $Ω$ and $BC$ ) and it must be in the same angle from $M$ , it must be the case that $ω → ω$ under the inversion. Consequently, it must mean that $D ↔ E$ and thus $M,D$ and $E$ are colinear. Another consequence of such argument is that the length of the tangents from $M$ to $ω$ are equal to $M B = M C$ .

Proof.

This is also quite a trivial statement, noticing from the previous statement that $CM 2 = M X ⋅AM$ and $CM 2 = M D ⋅M E$ we can conclude that $M X ⋅M A = M D ⋅M E$ which by the power of the point $M$ concludes that $AXDE$ is cyclic. Now, all that is left to notice is that, $∠ELD = ∠EDB$ due to $DB$ touching $ω$ and $∠EDB = ∠EAX$ due to $AXED$ being cyclic, thus $∠IAE = ∠ELD$ and $AILE$ is cyclic. ■

It is a bit interesting to see the behaviour of $(AILE )$ as one moves $I$ along $AM$ . When $I = X$ we get an already proven statement that $AXDE$ is cyclic and when $I = A$ we see that $(AILE )$ touches $ED$ . However, there is a more important position of $I$ which has the following property.

Proof. This again is no less trivial than the last statement, simply notice that $M D ⋅M E = M C2 = M I2$ , thus $(IDE )$ touches $AM$ . Consequently,

∠IDE = ∠AIE = ∠ALE

(3)

due to $AILE$ being cyclic. Which implies that $AL$ touches $ω$ . ■

2 Sawayama’s and Verrier’s Lemma’s

Proof. Now it is time to utilize all the lemmas proven in the previous section. The first step is to extend $CI$ till the intersection with $Ω$ , let that intersection point be $P$ . Then, due to the Trillium theorem it is clear that $PI = AP = P B$ . This allows one to apply the last lemma to this configuration. Here $C$ is serves as the arbitrary point from the last lemma and due to $IP = AP = PB$ it must be that the intersection of $IK$ with $ω$ , let that point be $′ L$ , must be the tangency point from $C$ to $ω$ .

However, that tangency point is $L$ by definition, thus $L ′ = L$ and consequently $L$ , $I$ and $K$ are colinear. ■

Now consider what happens when one moves $X$ along $AB$ , specifically let $X = A$ . Then, Sawayama’s lemma will transform itself into Verrier’s lemma.

This statement has other proofs which do not involve Sawayama’s lemma, one of the most notable ones is the following which showcases the mechanism at play.

Proof. Let us intersect $TL$ and $T K$ with $Ω$ in points $M1$ and $M2$ , by the shooting lemma it must be that $M1$ and $M2$ are the middle’s of arcs $⌢ AB$ and $⌢ AC$ . Consequently they must lie on the lines $BI$ and $CI$ .

All that is left is to apply Pascal’s theorem for $M1AM2BT C$ and conclude that $L$ , $I$ and $K$ are colinear. ■

Proof. Notice that due to the shooting lemma and its consequences, it must be that $M A2 = M L ⋅M T 1 1 1$ and $M A2 = M K ⋅M T 2 2 2$ . This, means that,

pow M = M 2= M L⋅M T = pow M (A,0) 1 1 1 1 ω 1

(4)

2 pow(A,0)M2 = M 2 = M2K ⋅M2T = powωM2

(5)

Thus, it must be that $M1$ and $M2$ lie on the radical axis of $(A, 0)$ and $ω$ , in other words $M1M2$ is the radical axis of $(A,0)$ and $ω$ . ■

3 Thebault’s theorem

Proof.

Notice, due to Sawayama’s lemma it must be that $LK ∩ M N = I$ , where $M, N$ are the tangency points of $ω2$ with $AX$ and $XC$ and $K, L$ are the tangency points of $ω1$ with $XB$ and $XC$ . With this in mind I suggest looking at the problem from another perspective, $LN$ is the inner tangent between $ω1$ and $ω2$ and $M K$ is the outer tangent between $ω1$ and $ω2$ . One must prove that $M N ∩ LK ∈ O1O2$ .

As it turns out this statement is true for arbitrary circles $ω1$ and $ω2$ .

Lemma 3. Let $ω1$ and $ω2$ be arbitrary circles prove that if $LN$ is the inner tangent line between them and $M K$ is the outer, then $M N ∩ LK ∈ O1O2$ .

This is a wonderful statement to consider on its own. Proving this statement automatically proves Thebault’s theorem. Let us consider the homothepy center $P$ which transforms $ω 1$ to $ω 2$ (in other words the intersection of the two outer common tangents). Let $X$ and $Y$ be the points of tangency of the common tangent of $ω 1$ and $ω 2$ . Let $A$ and $B$ be the intersection of $NL$ with $M K$ and $XY$ respectively. Consider the triangle $△ABP$ , then $ω 1$ is its incircle and $ω 2$ is its excircle. Then, by the Iran lemma the projection of $B$ onto the bisector of $∠BP A$ must lie on both $M N$ and $LK$ .

In other words, the projection of $B$ onto the bisector of $∠AP B$ is $M N ∩ LK$ . However, the projection of $B$ onto the bisector of $∠AP B$ obviously is part of $O1O2$ . Thus, $M N ∩ LK ∈ O1O2$ and the lemma is proven, proving Thebault’s theorem. ■

MrMineev