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 3 of 8: The excenter's angles at B and C
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N=excenter opposite E:∠NBC=90∘−c2,∠NCB=90∘−a2N=\text{excenter opposite }E:\quad \angle NBC=90^\circ-\tfrac c2,\quad \angle NCB=90^\circ-\tfrac a2
Detailed analysis

Let NN be the excenter of triangle BCEBCE opposite EE; it lies on the external bisectors of ∠EBC=c\angle EBC=c and ∠ECB=a\angle ECB=a, which meet BCBC at angles ∠NBC=90∘−c2\angle NBC=90^\circ-\tfrac c2 and ∠NCB=90∘−a2\angle NCB=90^\circ-\tfrac a2 (external bisector is perpendicular to the internal one).