Let ABCD be a convex quadrilateral with right angles at A and B. Prove that AD < BC if and only if the measure of the angle at D is greater than the measure of the angle at C.

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- March 15th 2010, 07:01 PMspectralseaConvex quadrilateral proof
Let ABCD be a convex quadrilateral with right angles at A and B. Prove that AD < BC if and only if the measure of the angle at D is greater than the measure of the angle at C.

- March 16th 2010, 03:16 AMGrandad
Hello spectralseaSuppose that is the foot of the perpendicular from to (or produced, if necessary). (The attached diagram shows the case where meets between and .)

Let .

Then we have:

is a rectangle (all angles right-angles)Now suppose that is measured positive in the same sense as .

(opposite sides of rectangle)

(angle sum of quadrilateral )

and are supplementary angles

if and only if is acute

if and only if is acuteGrandad

if and only if - March 16th 2010, 03:20 AMsa-ri-ga-ma
- March 16th 2010, 09:24 AMspectralsea
I'm so sorry, I must've been way out of it when I posted last night -- I completely forgot to include the fact that I'm working in non-Euclidean geometry. Therefore, I'm unable to use any of the parallelism or angle sum of quadrilateral. This is what I have so far on my own:

1)Assume <D > <C. By trichotomy, AD=BC, AD>BC, or AD<BC.

- Assume AD=BC. Then quad.ABCD is Saccheri, and <C=<D, which is a contradiction.

2) Assume AD<BC. By trichotomy, <D > <C, <D=<C, or <D < <C.

Any ideas how to complete those? - March 16th 2010, 10:50 AMPlato
You need to do #2 first. If then there is a point such that is a Saccheri quadrilateral.

Form that we get .

Now use that part to do the converse.