1. ## Convex 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.

2. Hello spectralsea
Originally Posted by spectralsea
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.
Suppose that $P$ is the foot of the perpendicular from $D$ to $BC$ (or $BC$ produced, if necessary). (The attached diagram shows the case where $P$ meets $BC$ between $B$ and $C$.)

Let $\angle BCD = \theta$.

Then we have:
$ADPB$ is a rectangle (all angles right-angles)

$\Rightarrow AD = BP$ (opposite sides of rectangle)

$\angle ADC = 180^o - \theta$ (angle sum of quadrilateral $ABCD$)

$\Rightarrow \angle ADC$ and $\angle BCD$ are supplementary angles

$\Rightarrow \angle BCD < \angle ADC$ if and only if $\angle BCD$ is acute
Now suppose that $PC$ is measured positive in the same sense as $BP$.
$\Rightarrow BC = BP + PC = AD + PC$
But
$PC = DC\cos\theta$
$\Rightarrow PC>0$ if and only if $\theta$ is acute

$\Rightarrow BC > AD$ if and only if $\angle BCD < \angle ADC$

3. Originally Posted by spectralsea
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.
Draw a parallel line D to AB which cuts BC at P. Now ABPD is a rectangle. So it is evident that angle ADC > angle BCD and AD < BC.

4. 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?

5. You need to do #2 first. If $BC>AD$ then there is a point $B-E-C$ such that $ABED$ is a Saccheri quadrilateral.
Form that we get $\left( {\angle CDA} \right) > \left( {\angle EDA} \right) = \left( {\angle BED} \right) > \left( {\angle BCD} \right)$.

Now use that part to do the converse.