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 1 of 8: Name the arcs and read off the key angles
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∠ACB=a, ∠DBC=c, ∠EBC=c, ∠ECB=a, ∠BEC=b+d\angle ACB=a,\ \angle DBC=c,\ \angle EBC=c,\ \angle ECB=a,\ \angle BEC=b+d
Detailed analysis

Let the arcs AB,BC,CD,DAAB,BC,CD,DA (not containing the other two vertices) have measures 2a,2b,2c,2d2a,2b,2c,2d with a+b+c+d=180∘a+b+c+d=180^\circ. Inscribed angles give ∠ACB=a\angle ACB=a, ∠DBC=c\angle DBC=c, and since E=AC∩BDE=AC\cap BD, ray BE=BE= ray BDBD and ray CE=CE= ray CACA, so in triangle BCEBCE: ∠EBC=∠DBC=c\angle EBC=\angle DBC=c, ∠ECB=∠ACB=a\angle ECB=\angle ACB=a, and ∠BEC=180∘−a−c=b+d\angle BEC=180^\circ-a-c=b+d.