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

If in a triangle 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, that it measures half the length of the third side, to which, in fact, it is also parallel.

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 that, therefore, it also cuts the third side of the triangle and measures half the length of the side that is parallel to it.

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

Let's look at an example

Given $⊿ABC$

$DE∥AB$ $AD=CD$

To prove: $DE=2AB$

Solution: If a straight line cuts one of the sides of a triangle – given that$DE$ cuts the edge $AC$, and is parallel to another side of the triangle,

Given that: $DE∥AB$ it means that it is a midsegment and therefore, measures half the length of the side it is parallel to.

That is: $DE=2AB$

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