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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- Mar 15th 2010, 06:01 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.

- Mar 16th 2010, 02:16 AMGrandad
Hello spectralseaSuppose that $\displaystyle P$ is the foot of the perpendicular from $\displaystyle D$ to $\displaystyle BC$ (or $\displaystyle BC$ produced, if necessary). (The attached diagram shows the case where $\displaystyle P$ meets $\displaystyle BC$ between $\displaystyle B$ and $\displaystyle C$.)

Let $\displaystyle \angle BCD = \theta$.

Then we have:

$\displaystyle ADPB$ is a rectangle (all angles right-angles)Now suppose that $\displaystyle PC$ is measured positive in the same sense as $\displaystyle BP$.

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

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

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

$\displaystyle \Rightarrow \angle BCD < \angle ADC$ if and only if $\displaystyle \angle BCD$ is acute

$\displaystyle \Rightarrow BC = BP + PC = AD + PC$But

$\displaystyle PC = DC\cos\theta$$\displaystyle \Rightarrow PC>0$ if and only if $\displaystyle \theta$ is acuteGrandad

$\displaystyle \Rightarrow BC > AD$ if and only if $\displaystyle \angle BCD < \angle ADC$ - Mar 16th 2010, 02:20 AMsa-ri-ga-ma
- Mar 16th 2010, 08:24 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? - Mar 16th 2010, 09:50 AMPlato
You need to do #2 first. If $\displaystyle BC>AD$ then there is a point $\displaystyle B-E-C$ such that $\displaystyle ABED$ is a Saccheri quadrilateral.

Form that we get $\displaystyle \left( {\angle CDA} \right) > \left( {\angle EDA} \right) = \left( {\angle BED} \right) > \left( {\angle BCD} \right)$.

Now use that part to do the converse.