# Midline and Midsegment of Trapezoid and Triangle

**trapezoid(trapezium)**.

The parallel sides of the trapezoid, are called **bases** (AB and CD) and the ones that are not parallel are called **legs** (AD and BC).

If the legs are equal in length, the trapezoid is called **isosceles**.

DE and CF are **altitudes**.

#### Midline of Trapezoid

A line that joins the midpoints of the sides that are not parallel is called a **midline(or a midsegment)** of trapezoid.

The line MN is the midline of ABCD. And the segment MN is the midsegment of ABCD.

AM = MD

BN = NC

The midline of a trapezoid is parallel to its sides.

In our case - MN || AB || DC.

*Theorem 1:*

*Theorem 2:*

In other words:

$\overline{MN} = \frac{\overline{AB} + \overline{DC}}{2}$

#### Midsegment of a Triangle

The segment that joins the midpoints of two sides of a triangle is called a **midsegment** of a triangle.

It is parallel to the third side and its length is half as long as the third side.

**Theorem**: If a line segment crosses the middle of one side of a triangle and is parallel to another side of the same triangle, then this line segment halves the third side.

$\overline{AM} = \overline{MC}$ and $\overline{BN} = \overline{NC}$ =>

$MN || AB$

$\overline{MN} = \frac{\overline{AB}}{2}$

#### Application of the properties of the midsegments

Divide a segment into equal segments without measuring.

**Assignment:** Divide a given segment $\overline{AB}$ into 5 equal segments without measuring.

**Solution:**

Let p be an arbitrary ray with origin A and that does not lie on AB. We draw consecutively five equal segments on p.

$\overline{AA_1} = \overline{A_1A_2} = \overline{A_2A_3} = \overline{A_3A_4} = \overline{A_4A_5}$

We connect A_{5} with B and draw lines through A_{4}, A_{3}, A_{2} and A_{1} that are parallel to A_{5}B.

They cross AB at points B_{4}, B_{3}, B_{2} and B_{1}, respectively.
These points divide the segment $\overline{AB}$ into five equal segments.

Indeed, from the trapezoid BB_{3}A_{3}A_{5} we see that $\overline{BB_4} = \overline{B_4B_3}$.
By the same way, from the trapezoid B_{4}B_{2}A_{2}A_{4},
we obtain $\overline{B_4B_3} = \overline{B_3B_2}$

While from the trapezoid B_{3}B_{1}A_{1}A_{3},

$\overline{B_3B_2} = \overline{B_2B_1}$.

Then, from B_{2}AA_{2}, it follows that $\overline{B_2B_1} = \overline{B_1A}$. We finally get:

$\overline{AB_1} = \overline{B_1B_2} = \overline{B_2B_3} = \overline{B_3B_4} = \overline{B_4B}$

It is clear that if AB must be divided into another number of equal segments, we must project the same number of equal segments on p. Then we proceed the same way.