Jack D'Aurizio Proof of Sawayama's Lemma

Proof

By Casey's theorem, applied to A, B, C (as 1-point circles) and circle C₁ (all "touching" C₂), we have that:

AF · BC + AB · CE = AC · BE

that is equivalent to:

where a, b, c are the lengths of sides BC, AC, and AB, as usual.

Let L be the inetersection of BC with the bisector AI of angle at A. Consider now the six points A, I, L, D, E, F. In virtue of Menelaus' theorem and its converse, E, F, I are collinear if and only if in ΔADL,

(LE/DE) · (DF/AF) · (AI/IL) = 1

Which is equivalent to:

(LE/AF) · (AI/IL) = 1

because DE = DF. In virtue of the angle bisector theorem and the Van Obel theorem we have that:

AI/IL = (b + c)/a,

so, in order to prove the collinearity of I, E, F, suffice it to prove that:

Again by the angle bisector theorem, BL = ac / (b + c). Now, LE = BE - BL = BE - ac/(b + c), so that (1) implies (2) and we are done.

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