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.
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
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?