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 7 of 8: The same ratio reappears for N
Previous stepNext step
∠LBN=∠LCN=90∘+d2,NBNC=sin⁡∠NCBsin⁡∠NBC=cos⁡(a/2)cos⁡(c/2)=sin⁡∠BLNsin⁡∠CLN\angle LBN=\angle LCN=90^\circ+\tfrac d2,\qquad \frac{NB}{NC}=\frac{\sin\angle NCB}{\sin\angle NBC}=\frac{\cos(a/2)}{\cos(c/2)}=\frac{\sin\angle BLN}{\sin\angle CLN}
Detailed analysis

Adding the bisector angles from steps 2 and 3: ∠LBN=∠LBC+∠CBN=c+d2+(90∘−c2)=90∘+d2\angle LBN=\angle LBC+\angle CBN=\tfrac{c+d}2+(90^\circ-\tfrac c2)=90^\circ+\tfrac d2, and likewise ∠LCN=∠LCB+∠BCN=a+d2+(90∘−a2)=90∘+d2\angle LCN=\angle LCB+\angle BCN=\tfrac{a+d}2+(90^\circ-\tfrac a2)=90^\circ+\tfrac d2; so ∠LBN=∠LCN\angle LBN=\angle LCN. The law of sines in triangle NBCNBC gives NB/NC=sin⁡∠NCB/sin⁡∠NBC=cos⁡(a/2)/cos⁡(c/2)NB/NC=\sin\angle NCB/\sin\angle NBC=\cos(a/2)/\cos(c/2), matching the previous step's ratio; and the law of sines in triangles NBL,NCLNBL,NCL together with ∠LBN=∠LCN\angle LBN=\angle LCN gives NB/NC=sin⁡∠BLN/sin⁡∠CLNNB/NC=\sin\angle BLN/\sin\angle CLN.