Let ABC be a triangle with AH altitude. The point K is chosen on the segment AH as follows such that AH=3KH. Let O be the center of the circle circumscribed around by triangle ABC,M and N be the midpoints of AC and AB respectively. Lines KO and MN intersect at the point Z, a perpendicular to OK passing through point Z intersects lines AC and AB at points X and Y respectively. Prove that ∠XKY=∠CKB. geometrycircumcirclemidpointsaltitudeequal angles