Given a rectangle ABCD such that AB=b>2a=BC, let E be the midpoint of AD. On a line parallel to AB through point E, a point G is chosen such that the area of GCE is
(GCE)=21(ba3+ab)
Point H is the foot of the perpendicular from E to GD and a point I is taken on the diagonal AC such that the triangles ACE and AEI are similar. The lines BH and IE intersect at K and the lines CA and EH intersect at J. Prove that KJ⊥AB. geometryrectanglearea of a trianglesimilar trianglesperpendicular