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Thread: Convex quadrilateral proof

  1. #1
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    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.
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  2. #2
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    Hello spectralsea
    Quote Originally Posted by spectralsea View Post
    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
    Attached Thumbnails Attached Thumbnails Convex quadrilateral proof-untitled.jpg  
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  3. #3
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    Quote Originally Posted by spectralsea View Post
    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.
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  4. #4
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    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?
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  5. #5
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    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.
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