(Deep breath) I've got an isosceles triangle ABC. There's a line DE cutting across it which looks like it's parallel to the base BC, but we don't know for sure.
The question is, given that the diagonals BE and DC are of equal length, show that the quadrilateral BCDE is cyclic.
If I could prove that BCDE is an isosceles trapezoid, I'd be in business...but I haven't been able to do that to my own satisfaction. Is it true that a quadrilateral has two equal base angles and diagonals of equal length, it must be an isosceles trapezoid? If so, could someone please sketch out how you'd prove it? I'm sure I'm missing something obvious.