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 C1 tangent to AD at F, CD at E, and the circumcircle C2 of ΔABC at K be centered at P.Then the chord EF passes through the incenter I of ΔABC.
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 C2. Then MN||EF due to C1 and C2 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.
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. C2, 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.
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 C3 centered at M
and radius BM passes through I. This circle is also orthogonal to C1. Indeed,
∠BKE = ∠MAC = ∠MBE
so that the circumcircle of ΔBKE is tangent to BM at B. C3 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, MB = MJ (but also MB = MI) so that J = I
and we are finished.
