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Trapezoid ABCD circumscribing cirlce O http://www.hicksvillepublicschools.org/cms/lib2/NY01001760/Centricity/Domain/1129/trapezoidcircle2.bmp

Definition: Two segments are called commensurable if they have a common measure

Definition: A common measure of two segments is a third segment such that it is contained in each of the first two a whole number of times with no remainder

So I have trapezoid ABCD which circumscribes circle O. I've constructed the midline of the trapezoid, EF. I've also constructed perpendiculars from points B and C down to segment AD.

Now, I know that EI is half of AG since it is the midline of triangle ABG. Similar argument for FJ and HD. And IJ is congruent to OH.

BC is congruent to IJ since BCJI is a parallelogram (easily proven)

AG + GH + HD + DC + BC + BA = perimeter EI + IJ + JF = midline

2EI + IJ + 2JF + DC + IJ + BA = perimeter

2EI + 2IJ + 2JF + DC + BA = perimeter

2*midline + DC + BA = perimeter.

And then I get stuck. I'm not even certain if I'm approaching this question the correct way.

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  • $\begingroup$ In your last equation, midline should be $2\cdot$midline. $\endgroup$
    – TonyK
    Commented Oct 18, 2015 at 17:16

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Hint: as your trapezoid is circumscribed to a circle, the distance from a vertex to the two tangency points of the sides that meet at that vertex is the same. It follows that $AB+CD=AD+CB$.

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  • $\begingroup$ I can't believe I forgot that!! Thank you! $\endgroup$
    – analysischallenged
    Commented Oct 18, 2015 at 17:32

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