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.
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
But
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?