Consider a triangle ABC. Let S be a circumference in the interior of the triangle that is tangent to the sides BC, CA, AB at the points D, E, F respectively. In the exterior of the triangle we draw three circumferences SA, SB, SC. The circumference SA is tangent to BC at L and to the prolongation of the lines AB, AC at the points M, N respectively. The circumference SB is tangent to AC at E and to the prolongation of the line BC at P. The circumference SC is tangent to AB at F and to the prolongation of the line BC at Q. Show that the lines EP, FQ and AL meet at a point of the circumference S. geometrygeometric transformationhomothety