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Parallelogram Requirement (Posted on 2013-07-07) Difficulty: 3 of 5

Let ABCD be a convex quadrilateral.

Prove that ABCD is a parallelogram
if and only if
|AC|2 + |BD|2 = |AB|2 + |BC|2 + |CD|2 + |DA|2.

  Submitted by Bractals    
Rating: 3.0000 (1 votes)
Solution: (Hide)

Let PQ denote the vector from point P to point Q.
Let PQ·RS denote the dot product of vector PQ and RS
with PQ·PQ = |PQ|2 = |PQ|2.

|AB|2 + |BC|2 + |CD|2 + |DA|2 =

  AB·AB + BC·BC + CD·CD + DA·DA =

  AB·AB + BC·BC + CD·CD + DA·DA +
    2AB·(AB + BC + CD + DA) =

  AB·AB + 2AB·BC + BC·BC +
    AB·AB + 2AB·DA + DA·DA +
    AB·AB + 2AB·CD + CD·CD =

  AB·AB + 2AB·BC + BC·BC +
    BA·BA + 2BA·AD + AD·AD +
    AB·AB + 2AB·CD + CD·CD =

  (AB + BC)·(AB + BC) +
    (BA + AD)·(BA + AD) +
    (AB + CD)·(AB + CD) =

  AC·AC + BD·BD +
    (AB + CD)·(AB + CD) =

|AC|2 + |BD|2 + |AB + CD|2

Therefore,

|AB|2 + |BC|2 + |CD|2 + |DA|2 = |AC|2 + |BD|2

  |AB + CD|2 = 0 ⇔ AB + CD = 0AB = DC

ABCD is a parallelogram.

QED

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionProof. No picture.Jer2013-07-09 11:00:04
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