In geometrical figure problems we draw many line segments to break the figure to get results easily. For example, to find the perimeter of a semicircle, divide the circle by drawing a line segment passing through its center called diameter. Similarly when a line joining two midpoints of two sides of a triangle is called mid-segment. Usually composite figures are best analyzed by dividing them into smaller figures. In this section will study about definition, properties and some theorems on triangle mid-segment

## Definition

In a triangle when a line segment is drawn joining the midpoint of one side of the triangle to the midpoint of another side of the triangle, then that line segment drawn is parallel to the third side of the triangle. The length of the line segment joining the middle points of the two sides of a triangle is also half the length of the third side of the triangle. This is known as mid segment of a triangle. In a triangle three such mid segments are possible to be drawn as a triangle has three sides.

## Properties

The properties of mid segments of a triangle are as follows:

• In a triangle a line segment connecting the middle point of two sides of triangle is parallel to third side of that triangle.

• In a triangle the length of line segment connecting the middle points of two sides of a triangle is half the length of the third side of the triangle.

• As a triangle is a geometric figure having three sides, we are able to draw three mid segments connecting the middle points of pair of sides.

## Theorem

Line segment joining mid points of any two sides of a triangle is parallel to the third side and has a length half of the length of the third side.

Proof:

$PM$ = $MR$ (given)

$PN$ = $NQ$ (given)

So, $PR$ = $PM + MR$ = $2\ PM$

$PQ$ = $PN + NQ$ = $2\ PN$

In triangle $PMN$ and triangle $PRQ$, angle $P$ is common

$\frac{PM}{PR}$ = $\frac{PN}{PQ}$ = $\frac{1}{2}$

As because the angle is same and the sides are in proportion we can say triangle $PMN$ and triangle $PRQ$ are similar triangles by side angle side postulate.

As $MN$ and $RQ$ are parts of the similar triangle $PMN$ and $PRQ$

Hence, $\frac{MN}{RQ}$ = $\frac{PM}{PR}$ = $\frac{PN}{PQ}$ = $\frac{1}{2}$ or $MN$ = $\frac{1}{2}$ $RQ$

Also, angle $PMN$ and angle $PRQ$ are corresponding angles of the similar triangle $PMN$ and $PRQ$ and we know that corresponding angles are always equal. So, angle $PMN$ = angle $PRQ$

The line segments $MN$ and $RQ$ are cut by the transversal $PR$ and the angle $PMN$ is equal to angle $PRQ$. Therefore, according to the parallel and transversal property we derive $MN$ is parallel to $RQ$. Hence, proved.

## Converse of Midsegment of a Triangle Theorem

The converse of mid segment of a triangle theorem states that if in a triangle a line is drawn through the midpoint of one side of a triangle and parallel to the other side of the triangle, then it intersects the third side of the triangle at its midpoint.

Proof:

We need to construct a line through $Q$ parallel to $RP$ and intersecting the extended line of $AB$ at $C$

Statement Reason
$AB // RQ$ Given
$RA //QC$ By construction
$ACQR$ is a parallelogram From statement $1$ and statement $2$
$RA$ = $QC$ Property of parallelogram opposite sides are equal
$RA$ = $PA$ Given
$PA$ = $QC$ From statement $4$ and statement $5$
Angle $APB$ = angle $BQC$ Considering triangle $APB$ and triangle $BQC$ alternate interior angles
as parallel line $AP$ and $QC$ cut by the transversal $PQ$
Angle $ABP$ = angle $QBC$ Vertically opposite angles
Triangle $APB$ is congruent to triangle
$QBC$
Angle angle side postulate of congruency
$PB$ = $BQ$ By corresponding parts of congruent triangles

Formula:

If $D$ is the midpoint of the side $AC$ and $E$ is the midpoint of the side $AB$ of the triangle $ABC$, then the line $DE$ is parallel to the third side $BC$ of the triangle $ABC$ and also half of it. The formula representing the theorem is $DE // CB$ and $DE$ = $\frac{1}{2}$ $CB$.

## Constructing the Midsegment of a Triangle

### Steps to construct the mid-segment of a triangle are as follows:

To construct the midpoint on any one of the side say $PR$ of the triangle $PQR$ drawn above, we place the compass needle at $P$ and draw two arcs one above and one below the side $PR$. Similarly, keeping the width of the compass unchanged we place the compass needle at $R$ and draw two arcs above and below the side $PR$. These two arcs intersect each other. Draw a line joining the two points of intersection. This line cuts the line $PR$ at its midpoint. The line becomes the bisector of the side $PR$. Let the midpoint be $M$. Following the same steps we draw the other two midpoints of the sides $PQ$ and $QR$ and mark them $N$ and $O$. Next on joining the pair of midpoints we get each of the midsegments $MN,\ NO$ and $MO$.

## Examples

Example 1:

If $DE$ is the length of the midsegment of the triangle $ABC$ is $7$ units, find $BC$.

Solution:

As per the midsegment theorem of triangle, length of midsegment is always parallel and half of the third side.

So, $DE // BC$ and $DE$ = $\frac{1}{2}$ $BC$.

Therefore, substituting the value of $DE$ = $7$ units

$7$ = $\frac{1}{2}$ $BC$

Multiplying both sides with $2$,

$7 \times 2$ = $\frac{1}{2}$ $\times\ 2\ \times\ BC$

$BC$ = $14$ units
Example 2:

In the diagram below $M$ is the midpoint of side $PQ$ and $N$ is the midpoint of side $PR$. Find the values of $x$ and $y$.

Solution:

As $M$ is the midpoint of $PQ$ and $N$ is the midpoint of $PR$, line segment $MN$ is the midsegment of the triangle $PQR$. So, it is parallel and half of the third side $QR$.

$PM$ = $MQ$

$2y + 6$ = $26$

$2y$ = $20$

$y$ = $10$

Also, $15x$ = $\frac{1}{2}$ $(20x + 10)$

$15x$ = $10x + 5$

$5x$ = $5$

$x$ = $1$

Therefore, the values of $x$ and $y$ are $1$ and 10.