MathLabs

Problem 3

Let ABCDABCD be a cyclic convex quadrilateral and Γ\Gamma be its circumcircle. Let EE be the intersection of the diagonals ACAC and BDBD, let LL be the center of the circle tangent to sides ABAB, BCBC, and CDCD, and let MM be the midpoint of the arc BCBC of Γ\Gamma not containing AA and DD. Prove that the excenter of triangle BCEBCE opposite EE lies on the line LMLM.
Step 2 of 8: L bisects the angles at B and C
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∠LBC=c+d2,∠LCB=a+d2\angle LBC=\tfrac{c+d}2,\quad \angle LCB=\tfrac{a+d}2
Detailed analysis

A circle tangent to lines AB,BCAB,BC has its center on the internal bisector of ∠ABC\angle ABC; tangent also to CDCD, its center lies on the internal bisector of ∠BCD\angle BCD too. Since ∠ABC=c+d\angle ABC=c+d and ∠BCD=a+d\angle BCD=a+d (each subtending the arc not containing the angle's own vertex), ∠LBC=c+d2\angle LBC=\tfrac{c+d}2 and ∠LCB=a+d2\angle LCB=\tfrac{a+d}2.