The Midsegment of a Trapezium

The Midsegment of a Trapezium — Moosa Academy

Join the midpoints of a trapezium's two legs and the segment you draw is not just any line across the middle. It runs parallel to both bases, and its length is exactly the average of theirs — one short formula that answers most questions on the topic.

Theorem The midsegment theorem

The midsegment of a trapezium is parallel to both bases, and its length is half their sum.

BE

is the midsegment of trapezium

ACDF

, then:

AF \parallel BE \parallel CD

BE = \dfrac{1}{2}(AF + CD)

The ticks mark the two halves of each leg. The midsegment must join the midpoints — a line drawn anywhere else across the trapezium has no such property.

Concept It is an average

Half the sum of two numbers is their average, so the midsegment always lies between the two bases in length — longer than the short one, shorter than the long one.

That gives a free check on any answer. If your midsegment comes out longer than both bases, or shorter than both, the arithmetic has gone wrong somewhere.

Example Finding the midsegment

In trapezium $ACDF$ , $BE$ is the midsegment. Given $AC = 14$ and $FD = 22$ , find $BE$ .

Apply the theorem:

BE = \dfrac{1}{2}(AC + FD)

Substitute:

BE = \dfrac{1}{2}(14 + 22)

Add inside the bracket:

BE = \dfrac{1}{2}(36)

$BE = 18$

The check holds: $18$ sits neatly between $14$ and $22$ .

Example Working backwards to a base

In trapezium $ACDF$ the midsegment $BE = 26$ . Given $AC = 3x - 4$ and $FD = 5x + 2$ , find $x$ and both bases.

Apply the theorem:

26 = \dfrac{1}{2}\bigl[(3x - 4) + (5x + 2)\bigr]

Double both sides:

52 = 8x - 2

Add

2

54 = 8x

Divide:

x = 6.75

AC = 3(6.75) - 4 = 16.25

FD = 5(6.75) + 2 = 35.75

$x = 6.75$ , with bases $16.25$ and $35.75$

Check by substituting back: $\dfrac{1}{2}(16.25 + 35.75) = \dfrac{52}{2} = 26$ , the midsegment given. Doubling both sides at the start is what clears the fraction and keeps the algebra simple.

Summary

The midsegment joins the midpoints of the two legs — not any other pair of points.
It is parallel to both bases at once.
Its length is half the sum of the two bases.
Half a sum is an average, so the midsegment always lies between the two base lengths.
To find a base from the midsegment, double both sides first to clear the fraction.
Always substitute your answer back — the average check catches most slips instantly.