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Trapezoid Median

Date: 14 Aug 1995 16:13:12 -0400
From: Damien Mascord
Subject: Proof for a geo q

Question: PQRS is a trapezium with PQ||SR.  If A and B are mid-points
of SP and RQ respectively prove that

a) AB is parallel to PQ and SR,

b) Ab=1/2(PQ + SR)

    /           \
  /               \

Date: 15 Aug 1995 17:22:44 -0400
From: Dr. Ken
Subject: Re: Proof for a geo q

Hello there!

To do this problem, you kind of have to work backwards: instead of saying
that A and B are midpoints of PS and RQ, say that A is the midpoint of PS
and that line L is the line through A which is parallel to P and S.  Since
we're in Euclidean Geometry, there's only one such parallel line.  Let C be
the point at which this line intersects QR (which it does, again because
we're in Euclidean Geometry and not something else funky).

    /           \
  /               \

Now since we have three parallel lines (SR, AC, and PQ) with two
transversals (PS and QR), the ratio of SA to AP must be the same as the
ratio of RC to CQ.  Lo and behold, that ratio is 1:1.  So C is the midpoint
of RQ.  Thus the lines AC and AB (as you originally defined it) are the same
line.  So AB is parallel to PQ and SR.

To get the second part of the problem, drop all these perpendiculars:

    /|         |\
  /| |         | |\
   D E         F G

Then since all the things that look like rectangles really are, you have the
following equations:  

AB = PQ + DE + FG
SD = AH (from congruent triangles) = DE
GR = JB (from congruent triangles) = FG.

So from the last two equations, D and G are the midpoints of SE and FR,
respectively.  So 

DE = SE/2, and 
FG = FR/2.  So from the very first equation, 
AB = PQ + SE/2 + FG/2.  

And since SR = PQ + SE + FG, you can probably see that AB is the average of
PQ and SR.  

Anyway, let us know if you didn't follow something, or if you think we
messed up somewhere.

Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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