Midsegment

The midsegment is a segment that connects the midpoints of 2 sides.
It's very simple to remember the meaning of this term since the word "middle" already tells us that it is about the midpoint, so when we come across the concept of "midsegment" we'll remember that it connects the midpoints of two sides.
We're here to teach you everything you need to know about the midsegment, from the proof to the wonderful properties of the segment that will help us solve exercises.
First, we'll talk about the midsegment of a triangle and then we'll move on to the midsegment of a trapezoid.

The midsegment of a triangle crosses through the midpoint of the two sides from which it extends, but, beyond this, it has two remarkable properties that we can utilize after proving that this segment is, indeed, a midsegment of the triangle.

The midsegment of a triangle is half the length of the third side
and is also parallel to it.

If $AD=CD$
$AE=BE$
then $2DE=CB$
$DE∥CB$

The theorem discusses the properties of the midsegment and its definition.
We can determine that we are looking at a midsegment of a triangle if at least one of the following conditions is met:

• If in a triangle there is a straight line extending from the midpoint of one side to the midpoint of another side, we can determine that it is a midsegment and therefore, measures half the length of the third side, to which it is also parallel.

That is, if we know that:

$AE=BE$
$AD=CD$

Then, we can determine that:
$DE$ is a midsegment of a triangle and, consequently,

$2DE=BC$
$DE\parallel BC$

• If a straight line cuts one of the sides of a triangle and it is parallel to another side of the triangle, it means that it is a midsegment and, therefore, also cuts the third side of the triangle in half the length of the side that is parallel to it.

That is, if we know that:

$AD=CD$
and also $DE\parallel BC$

Then, we can determine that:
$DE$ is a midsegment of a triangle and, consequently,
$AE=BE$
and also $2DE=BC$

• If in a triangle there is a segment whose endpoints are on two of its sides, it measures half the length of the third side and is parallel to it, we can determine that such a segment is a midsegment and, therefore, cuts the sides it touches right in the middle.

That is, if we know that:

$2DE=BC$
and also $DE\parallel BC$

Then, we can determine that:
$DE$ is a midsegment of a triangle and, consequently,
$AE=BE$
and also $AD=CD$

Some notes for guaranteed victory

• We realize that it is easy to remember that a midsegment is a segment that goes from the midpoint of one side to the midpoint of another side since the word itself reveals it. But pay attention! The term "midsegment" is not used in every case; sometimes the term "median" is used to describe the line that cuts a side through the midpoint.
  So, when you come across the word "median," remember that you may need to look for a triangle's midsegment.
• There are figures with medians by definition. For example, the diagonals of a parallelogram intersect in the middle; thus, they are medians. Therefore, if we draw a segment from the intersection point of the diagonals to the edge of the parallelogram, this will be a midsegment in the corresponding triangle formed by the diagonal.

The mid-segment of a trapezoid has properties very similar to those of the mid-segment of a triangle... It makes sense since, after all, we're still talking about the mid-segment.

The midsegment of a trapezoid bisects the two non-parallel sides it emerges from and is parallel to both bases of the trapezoid, as well as being half the length of these.
Notice, as we have already mentioned, its properties are similar to those of the midsegment of a triangle.
The two expressions that you should remember are: parallel and measures half.
But, don't be mistaken, in the trapezoid the midsegment measures half the length of the bases - that is, half the length of both bases combined.
You will be able to use these properties after proving that there is a midsegment in the trapezoid.
Let's look at the properties of the midsegment illustrated:

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

The Midsegment Theorem in a trapezoid is about the properties of midsegments.
If at least one of the following conditions is met, we can determine that it is a midsegment in a trapezoid:

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

That is, if we know that:
$AE=DE$
and also
$BF=CF$
Then we can determine that:
$EF$ is a midsegment of the trapezoid
Therefore:
$AB∥EF∥DC$

$EF=\frac{AB+DC}{2}$

If there is a straight line in a trapezoid that comes off one side and is parallel to one of the bases of the trapezoid, we can determine that it is a midsegment and, therefore, 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.

That is, if we know that:

$AE=DE$
and also
$AB∥EF$
Then we can determine that:
$EF$ is a midsegment of the trapezoid and, as a result:
$AB∥EF∥DC$
$BF=CF$
$EF=\frac{AB+DC}{2}$