Re: Question about trapezoids

[Jason Fahy]

Poor phrasing up there, let me try again:

Is it true that IF a quadrilateral has two equal base angles and diagonals of equal length, it must be an isosceles trapezoid? If so, etc, etc...

[Jason Fahy]

(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.

Thanks,
JF

[Walter Whiteley]

The problem seems to come down to special cases of SSA.

When you speak about 'diagonals' there is am implication that these
are longer than the missing third side, and that the 'diagonals'
cross while the missing third sides do not cross.

If this is true, then you are in a subcase where SSA gives the
necessary congruence.

Walter Whiteley

[Ed Wall]

This is sort of what is being said below: Okay, draw your isosceles
triangle. Chose a length a little longer than the base and using, for
example, B as a center draw a circle intersecting AC at two points.
Label E the one nearer to C. Taking C as a center and using the same
length do the same for BA; however, label D the one nearer to A. Now
DC and BE are certainly equal, but is BCED an isosceles trapezoid (I
notice you are talking about BCDE so perhaps I misunderstand)?

Ed Wall

[Jason Fahy]

(I notice you are talking about BCDE so perhaps I misunderstand)?

JF> No, that was just sloppy labeling on my part. I do angles correctly but for polygons I have a bad habit of alphabetizing the vertices instead of giving them in the proper order. :/

[Ara M Jamboulian]

BCED is not necessarily isosceles.
So you need to prove cyclic using some other method.

[Walter Whiteley]

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).

Walter

[Peter Chang]

In fact the theorem we are to approve is totally wrong. We can disprove it easily.

In an isosceles ABC, Dram line segment BE first. When drawing line CD , you will find you can draw two lines that are equal to BE.

Draw a circle with radius CD, and C as its center. As generally CD is not perpendicular to AB, the circle will intersect line AB twice, at D and another point, say D'. CD = CD'.

Now connect E and D, E and D'. If ED' is parallel to BC, then quadrilateral BCED' is cyclic, and BCED is not. If ED is parallel to BC, then quadrilateral BCED is cyclic, and BCED' is not.

So now consider the following questions:
(1) In isosceles ABC, DC = BE, is quadrilateral ABCD cyclic?
(2) In isosceles ABC, D'C = BE, is quadrilateral ABCD' cyclic?

Note that to make the theorem true, we need to add another condition: CD and EB are perpendicular to AB and AC respectively.