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 5 of 8: The key matching of angles
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∠MBL=90∘−a2=∠NCB,∠MCL=90∘−c2=∠NBC\angle MBL=90^\circ-\tfrac a2=\angle NCB,\qquad \angle MCL=90^\circ-\tfrac c2=\angle NBC
Detailed analysis

Adding the previous two steps' angles at BB: ∠MBL=∠MBC+∠LBC=b2+c+d2=b+c+d2=90∘−a2\angle MBL=\angle MBC+\angle LBC=\tfrac b2+\tfrac{c+d}2=\tfrac{b+c+d}2=90^\circ-\tfrac a2, which is exactly ∠NCB\angle NCB. Symmetrically, ∠MCL=∠MCB+∠LCB=b2+a+d2=90∘−c2=∠NBC\angle MCL=\angle MCB+\angle LCB=\tfrac b2+\tfrac{a+d}2=90^\circ-\tfrac c2=\angle NBC.