Midsegment of a trapezoid

The midsegment of a trapezoid divides into two equal parts the two sides from which it extends and, in addition, is parallel to both bases of the trapezoid and measures half the length of these.
Let's see the properties of the midsegment of a trapezoid in the following illustration:

1- Let's see the properties of the midsegment in an illustration

If $EF$ Midsegment
then:
$AE=DE$
$BF=CF$
$AB∥EF∥DC$
$EF=\frac{AB+DC}{2}$

Detailed explanation

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Test yourself on midsegment of a trapezoid!

Below is an isosceles trapezium.

EF is parallel to the base of the trapezium.

True or false: EF is the midsection of the trapezoid.

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Proof of the Midsegment of a Trapezoid

We can demonstrate that there is a midsegment in a trapezoid if at least one of the following conditions is met:

If in a trapezoid there is a straight line that extends from the midpoint of one side to the midpoint of another side, we can determine that it is a midsegment and, therefore, it is parallel to both bases of the trapezoid and measures half the length of these.
If in a trapezoid there is a straight line that comes out of one side and is parallel to one of the bases of the trapezoid, we can determine that it is a midsegment and, therefore, it is parallel to both bases of the trapezoid, measures half the length of these two and also cuts the second side it touches in half.

Let's look at an example

Data:

$AE=DE$
$AB‖EF$

To prove:
$BF=FC$

Solution:
According to the second rule of the theorem,
if in the trapezoid, a straight line $EF$ comes out from the midpoint of one of the sides - we know that: $AE=DE$
and is also parallel to one of the bases of the trapezoid – Given that: $AB‖EF$
we can determine that it is the mid-segment of the trapezoid.
Consequently, we can determine that it also cuts through the midpoint of the other side it touches.
That is to say:
$BF=FC$