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Math Help - Diagonals in a quadrilateral

  1. #1
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    Diagonals in a quadrilateral

    In a converx quadrilateral ABCD, points M and N are midpoints of sides AB and CD, and diagonals meet at E. Show that the line containing the bisector of angle BEC is perpendicular to line MN iff AC=BD.

    My repeated attempts at this problem prove vain, your help would be greatly appreciated. I attach a picture which has, however, not helped me in any way.




    Diagonals in a quadrilateral-pro3.png
    Last edited by CaptainBlack; October 2nd 2010 at 11:03 PM.
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  2. #2
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    Here's a vector proof that seems to work.

    Denoting the bisector of BEC as EL, the scalar product of \underline{EL} with unit vectors in the directions of EC and EB will be equal, so,

    \displaystyle\frac{\underline{EC}}{\mid \underline{EC}\mid}.\underline{EL}=\frac{\underlin  e{EB}}{\mid\underline{EB}\mid}.\underline{EL}\ \ \dots\dots(1)

    Now,

    \displaystyle\underline{MN}=\underline{MA}+\underl  ine{AC}+\underline{CN}

    and

    \displaystyle\underline{MN}=\underline{MB}+\underl  ine{BD}+\underline{DN},

    so adding, and dividing by 2,

    \underline{MN}=\frac{1}{2}(\underline{AC}+\underli  ne{BD})=\frac{1}{2}(\lambda\underline{EC}-\mu\underline{EB}), for some constants \lambda and \mu.

    If MN and EL are to be at right angles to each other, their scalar product will have to equal zero so, using (1),

    \lambda will have to be some multiple of 1/\mid\underline{EC}\mid and \mu will have to be the same multiple of 1/\mid\underline{EB}\mid implying in turn that \underline{AC} and \underline{BD} are equal in magnitude.
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  3. #3
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    Hello atreyyu
    Quote Originally Posted by atreyyu View Post
    In a converx quadrilateral ABCD, points M and N are midpoints of sides AB and CD, and diagonals meet at E. Show that the line containing the bisector of angle BEC is perpendicular to line MN iff AC=BD.

    My repeated attempts at this problem prove vain, your help would be greatly appreciated. I attach a picture which has, however, not helped me in any way.






    Click image for larger version. 

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    On your diagram, construct the bisector of \angle AEB, meeting AB and CD at P and Q, respectively. Then, since this bisector is perpendicular to the bisector of \angle BEC, we need to prove that PQ \parallel MN iff AC = BD.

    Let the points of intersection of MN with AC and BD be R and S respectively, and suppose that:
    AM = MB = m

    DN=NC=n

    AE = b

    BE = a

    EC=d

    ED =c
    Then, using the angle bisector theorem on \triangle AEB
    \dfrac{AP}{AE}=\dfrac{PB}{EB}

    \Rightarrow AP = \dfrac{b}{a}PB

    \Rightarrow AP +PB = 2m = \left(\dfrac{a+b}{b}\right)PB

    \Rightarrow PB = \dfrac{2ma}{a+b}

    and AP = \dfrac{2mb}{a+b}

    Similarly
    DQ = \dfrac{2nc}{c+d}

    QC= \dfrac{2nd}{c+d}

    Now we'll assume that PQ \parallel MN and prove that AC =BD. I'll leave you to prove it the other way around.

    So, from \triangle DEQ, NS\parallel QE:
    \dfrac{DN}{DQ}=\dfrac{DS}{DE}

    \Rightarrow \dfrac{n(c+d)}{2nc}=\dfrac{DS}{c}

    \Rightarrow DS = \frac12(c+d)

    Again, in a similar way, from \triangle BMS:
    BS = \frac12(a+b)

    But BS + SD = BD = a+c
    \Rightarrow \frac12(a+b) +\frac12(c+d)=a+c

    \Rightarrow a+ b+c+ d = 2a+2c

    \Rightarrow b+d = a+c

    \Rightarrow AC = BD
    Grandad
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    Thank you both big time
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