Chalenging problem?

Question

Consider five points A, B, C, D and E such that ABCD is a parallelogram and BCED is a cyclic quadrilateral. Let \ell be a line passing through A. Suppose that \ell intersects the interior of the segment DC at F and intersects line BC at G. Suppose also that EF=EG=EC. Prove that \ell is the bisector of angle DAB.

Answer 1

I don't know if this is what you're looking for, but based on the way you labelled the figures, ABCD is the parallelogram labelled in CW order. BCED is the cyclic quadrilateral labelled in CW order.

This means that BC and CD are adjacent sides of the parallogram which intersect at C. If the line intersects CD as well as BC, then it must intersect at C which makes C, F and G the same point. Thus EC = EF = EG.

This would also make the line the diagonal of the parallelogram, which can be easily shown to bisect angle DAB. See below.

Call angle CAB = x = ACB
Thus ABC = 180 - 2x
Also since ABC + BAD = 180
BAD = 2x
Thus AC = AF = AG bisects angle BAD.

Note that the diagonal is the only line that bisects an angle of a parallelogram.

But even if for some mysterious reason, we let C, F and G be distinct points, such that G is on BC extended. Then since each is equidistance from E, we can draw a circle with center E through C, F and G.
CG // BC
If we draw a line through F parallel to CG, then we can form a parallelogram FCGH, which is similar to ABCD. Since line AFG is the diagonal of this parallelogram, it bisects the angle CFH which is the same angle as DAB.

Answer 2

Right on Dr. D.

But I cant help thinking that he/she may have stated the problem incorrectly.