> (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 > d prove it? I'm sure I'm missing something obvious. > > Thanks, > JF BCED is not necessarily isosceles. So you need to prove cyclic using some other method.