# Y. Sawayama's Lemma: What Is It About?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny

The applet purports to illustrate Y. Sawayama's lemma:

Through vertex A of ΔABC a straight line AD is drawn with D on BC. Let circle *C _{1}* tangent to AD at F, CD at E, and the circumcircle

*C*of ΔABC at K be centered at P.Then the chord EF passes through the incenter I of ΔABC.

_{2}Sawayama's paper has been discovered by J.-L. Ayme who used the lemma to directly solve Thébault's Problem. The proof by Ayme is a slight modification of that by Sawayama. Along the way, Ayme corrects a logical gap in the original proof.

Let M and N be the intersections of KE and KF with the circumcircle

*C*. Then MN||EF due to_{2}*C*and_{1}*C*being homothetic at K. M is the midpoint of the arc BC not including K. Therefore, AM is the bisector at A and, thus, contains the incenter I._{2}Let J be the intersection of AM and EF. Consider the configuration of two lines AM and KN and the coaxal system of circles through A and K. One of the circles, viz.

*C*, cuts the chord MN. Another circle - that through F - cuts a parallel chord, bound to lie on EF. This implies that quadrilateral AFJK is cyclic._{2}Apply now Miquel's Pivot Theorem to ΔAFJ with F on AF, E on FJ, and J on AJ. K is the pivot point common to the three circles. Circle EJK is tangent to AJ (same as AM) at J.

Circle

*C*centered at M and radius BM passes through I. This circle is also orthogonal to_{3}*C*. Indeed,_{1}∠BKE = ∠MAC = ∠MBE

so that the circumcircle of ΔBKE is tangent to BM at B.

*C*is orthogonal to the latter circle and, since M lies on EK, to all circles through K and E, in particular, to the circle EJK. Therefore,_{3}MB = MJ (but alsoMB = MI) so thatJ = I

and we are finished.

### Reference

- J.-L. Ayme,
__Sawayama and Thébault's Theorem__,*Forum Geometricorum*, v 3 (2003), 225-229. - Y. Sawayama,
__A New Geometrical Proposition__,*Amer. Math. Monthly*, 12 (1905) 222-224.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

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