The Midsegment of a Trapezoid

Submitted by Isoscel on Sat, 11/17/2007 - 20:13.

In a trapezoid ABCD (with $BC \parallel AD$ ) if M is the middle of AB and N is the middle of CD then MN is called the midsegment of the trapezoid.

MN is parallel with BC and DA and is equal with the arithmetic mean of BC and DA:
$MN=\frac{BC+DA}{2}$
(in a convex trapezoid)

Proof:

Let P be the middle of the segment AC.

In the triangle ABC, M is the middle of AB and P is the middle of CA so MP is the midsegment of the triangle ABC. This means that $MP=\frac{BC}{2}$ and $MP \parallel BC$ .

In the triangle ACD, P is the middle of AC and N is the middle of CD so PN is the midsegment of the triangle ACD. This means that $PN=\frac{AD}{2}$ and $PN \parallel AD$ .

$MP \parallel BC$ and $BC \parallel AD$ so $MP \parallel AD$ . But also $PN \parallel AD$ so M, P and N are collinear points. Because ABCD is convex and P is the middle of AC, and M, N are on the sides of the trapezoid, P must be between M and N on the line. So $MN = MP + PN$ .

$MN = MP + PN = MP=\frac{BC}{2}+\frac{AD}{2}=\frac{BC+AD}{2}$

M, N and P are collinear, and $MP \parallel BC$ and $PN \parallel AD$ .
So $MN \parallel BC$ and $MN \parallel AD$ .

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The Midsegment of a Trapezoid