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 8 of 8: Conclude via the uniqueness lemma for sines
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∠BLM=∠BLN ⇒ L,M,N collinear\angle BLM=\angle BLN\ \Rightarrow\ L,M,N\ \text{collinear}
Detailed analysis

Both MM and NN lie inside angle ∠BLC\angle BLC, so ∠BLM+∠MLC=∠BLC=∠BLN+∠NLC\angle BLM+\angle MLC=\angle BLC=\angle BLN+\angle NLC. Combined with sin⁡∠BLM/sin⁡∠MLC=sin⁡∠BLN/sin⁡∠NLC\sin\angle BLM/\sin\angle MLC=\sin\angle BLN/\sin\angle NLC (steps 6-7), the elementary fact that two pairs of positive angles with equal sum less than 180∘180^\circ and equal sine ratio must be respectively equal forces ∠BLM=∠BLN\angle BLM=\angle BLN. Hence rays LMLM and LNLN coincide, so NN lies on line LMLM, as required.