Alternatively - given the correct observation below that there are a couple of options for the second diagonal, given the first one - you actually find that the points may not be cyclic!
At least I am looking at a counterexample on my computer!
So you may want to know something more about the length of the diagonals in comparison to say the base of the triangle (so that the diagonals become unique).
On 18-Jan-07, at 10:08 AM, Ara M Jamboulian wrote:
> (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.