MathDB
Line through circumcentres forms an isosceles triangle

Source: Baltic Way 2005

December 28, 2010
geometrygeometric transformationrotationcircumcirclerhombuspower of a pointradical axis

Problem Statement

Let the points DD and EE lie on the sides BCBC and ACAC, respectively, of the triangle ABCABC, satisfying BD=AEBD=AE. The line joining the circumcentres of the triangles ADCADC and BECBEC meets the lines ACAC and BCBC at KK and LL, respectively. Prove that KC=LCKC=LC.
Back to ProblemsView on AoPS