Let O be the circumcenter of triangle ABC, M, D midpoints of BC, CA, E projection of D onto AB. Prove that KE=KC.
Proof:
Let BK intersect the perpendicular at C to BC at the point F, then OMCF is a rectangle, i.e. cyclic; also OMCD is cyclic, having 2 opposite angles of 90 degs each. Consequently O, M, C, F, D are concyclic points and <FDC=<FMC=<OCM=90-A=<ADE, so F-D-E are collinear, BCFE is cyclic, its circumcenter being K, midpopint of OM, done.
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