Difference between revisions of "Sparrow’s lemmas"

Revision as of 14:23, 16 September 2025

Sparrow’s lemmas have been known to Russian Olympiad participants since at least 2016.

Let triangle $ABC$ with circumcircle $\Omega$ and points $D$ and $E$ on the sides $AB$ and $AC,$ respectively be given.

Let $K \in \Omega$ be the midpoint of the arc $BC$ which contain the point $A.$

Prove that $BD = CE$ iff points $A, D, E,$ and $K$ are concyclic.

Proof

$BK = CK, \angle ABK = \angle ACK.$

Let $BD = CE \implies \triangle BKD = \triangle CKE \implies$

$\angle KDA = \angle KEA \implies A, D, E,$ and $K$ are concyclic.

Let $A, D, E,$ and $K$ are concyclic $\implies \angle KDA = \angle KEA \implies$

$\angle BKD = \angle CKE \implies \triangle BKD = \triangle CKE \implies BD = CE.$

@@ Line 1: / Line 1: @@
 Sparrow’s lemmas have been known to Russian Olympiad participants since at least 2016.
-==Lemma 1==
+==Sparrow's Lemma 1==
 [[File:Locus.png|300px|right]]
 Let triangle <math>ABC</math> with circumcircle <math>\Omega</math> and points <math>D</math> and <math>E</math> on the sides <math>AB</math> and <math>AC,</math> respectively be given.