Sparrow’s lemmas

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

Sparrow's Lemma 1

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 $\implies \angle KDA = \angle KEA \implies$

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

Sparrow’s Lemma 1A

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

Let $M$ be the midpoint of $BC, I$ be the incenter.

Prove that $BD + CE = BC$ iff points $M, D, E,$ and $I$ are concyclic.

Proof

1. Let points $M, D, E,$ and $I$ are concyclic.

Denote $F \in BC$ such $BD = BF, \varphi = \angle ADI.$

So point $F$ is symmetric to $D$ with respect to $BI \implies \angle BDI = 180^\circ - \varphi, \angle IEC = \varphi.$ $\triangle BDI = \triangle BFI \implies DI = FI, \angle CFI = \varphi.$ $\angle DAI = \angle EAI \implies DI = EI = FI.$ $\triangle CIF = \triangle CIE \implies CE = CF \implies BD + CE = BC \blacksquare$ 2. Let $BD + CE = BC \implies$ there is point $F$ such that $BF = BD, CF = CE \implies$ $\triangle CIF = \triangle CIE, \triangle BIF = \triangle BID \implies$ $180^\circ = \angle BFI + \angle CFI = \angle BDI + \angle CEI = \angle ADI + \angle AEI \blacksquare$

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Sparrow’s lemmas

Sparrow's Lemma 1

Sparrow’s Lemma 1A