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 6 of 8: Sine rule in triangles MBL and MCL
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sin⁡∠BLMsin⁡∠CLM=sin⁡∠MBLsin⁡∠MCL=cos⁡(a/2)cos⁡(c/2)\frac{\sin\angle BLM}{\sin\angle CLM}=\frac{\sin\angle MBL}{\sin\angle MCL}=\frac{\cos(a/2)}{\cos(c/2)}
Detailed analysis

By the law of sines in triangles MBLMBL and MCLMCL (sharing side MLML), MBsin⁡∠MLB=MLsin⁡∠MBL\dfrac{MB}{\sin\angle MLB}=\dfrac{ML}{\sin\angle MBL} and MCsin⁡∠MLC=MLsin⁡∠MCL\dfrac{MC}{\sin\angle MLC}=\dfrac{ML}{\sin\angle MCL}. Since MB=MCMB=MC, dividing gives sin⁡∠BLMsin⁡∠CLM=sin⁡∠MBLsin⁡∠MCL=sin⁡(90∘−a/2)sin⁡(90∘−c/2)=cos⁡(a/2)cos⁡(c/2)\dfrac{\sin\angle BLM}{\sin\angle CLM}=\dfrac{\sin\angle MBL}{\sin\angle MCL}=\dfrac{\sin(90^\circ-a/2)}{\sin(90^\circ-c/2)}=\dfrac{\cos(a/2)}{\cos(c/2)}.