Results 1 to 5 of 5

Math Help - Convex quadrilateral proof

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    14

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    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 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
    Grandad
    Attached Thumbnails Attached Thumbnails Convex quadrilateral proof-untitled.jpg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2010
    Posts
    14
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,391
    Thanks
    1476
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Regular Quadrilateral Proof
    Posted in the Geometry Forum
    Replies: 6
    Last Post: May 2nd 2011, 02:56 PM
  2. Convex Function Proof
    Posted in the Differential Geometry Forum
    Replies: 8
    Last Post: January 27th 2011, 02:29 PM
  3. Convex set proof
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: August 30th 2010, 06:58 PM
  4. Convex set proof
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: February 18th 2010, 10:29 AM
  5. Convex set proof
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: October 14th 2009, 03:20 AM

Search Tags


/mathhelpforum @mathhelpforum