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