Midsegment of a Triangle: Definition, Theorem, Formula, Examples

What Is the Midsegment of a Triangle?

The midsegment of a triangle is a line segment joining the midpoints of two sides of the triangle. A triangle has three midsegments.

Consider the triangle PQR in the given diagram. The points P, Q, and R are the three vertices of the triangle. PQ, QR, and PR are the three sides. Point A is the midpoint of side PR. Point B is the midpoint of side PQ. 

The line segment AB that joins the midpoints A and B. It is the midsegment of the triangle PQR.

Notice that AB is parallel to the side QR.

Midpoint is the middle or center point of a line segment that divides it into two equal parts. It is equidistant from both endpoints of a line segment. A triangle is a polygon with 3 sides, 3 angles, and 3 vertices.

Midsegment of a Triangle: Definition

The midsegment is a line segment that joins the midpoints of any two sides of a triangle. 

So, how many midsegments does a triangle have? The answer is 3.

In the triangle ABC shown below, the three midsegments are XY, YZ, and XZ.

Here, 

AB || XY

AC || YZ

BC || XZ

Properties of the Midsegment

• A triangle has three midsegments. 

DE, EF, DF are three midsegments of the triangle ABC.

• A midsegment joins the midpoints of two sides.

D is the midpoint of the side AB.

E is the midpoint of the side AC.

F is the midpoint of the side BC.

• The midsegment of a triangle is always parallel to the third side of the triangle.

DE || BC

EF || AB

DF || AC

• The length of the midsegment is half the length of the third side.

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

EF $=\frac{1}{2}$ AB

DF $=\frac{1}{2}$ AC

• The three midsegments of a triangle form a triangle similar to the original triangle. 

           DEF   ABC

• The smaller triangle formed by the midsegments is $\frac{1}{4}$th area of the original triangle and half the perimeter of the original triangle.

Area of $\mathrm{\Delta }\text{BDF}=\text{Area of}\mathrm{\Delta }\text{CEF}=\text{Area of}\mathrm{\Delta }ADE=\frac{1}{4}\times \mathrm{\Delta }\text{ABC}$

Perimeter of $\mathrm{\Delta }\text{BDF}=\text{Perimeter of}\mathrm{\Delta }\text{CEF}=\text{Perimeter of}\mathrm{\Delta }\text{ADE}=\frac{1}{2}\times \text{Perimeter of}\mathrm{\Delta }\text{ABC}$

Midsegment Theorem of a Triangle

According to the midsegment theorem:

• AB || RQ, AB $=\frac{1}{2}$ RQ
• AC || PQ, AC$=\frac{1}{2}$ PQ
• BC || PR, BC$=\frac{1}{2}$ PR

Proof of Midsegment Theorem of a Triangle

To prove: 

In the triangle PQR,

• AB || RQ

• AB $=\frac{1}{2}$ RQ

Construction: Extend the line AB from the point B. Construct a line QC parallel to AR, such that it meets the extended line AB at point C. 

A is the midpoint of PR. B is the midpoint of PQ.

In $\mathrm{\Delta }PAB$ and $\mathrm{\Delta }QCB$

$PB=BQ$ $\therefore$ B is the midpoint of PQ 

$\mathrm{\angle }APB=\mathrm{\angle }CQB$ $\therefore$ Alternate angles

$\mathrm{\angle }ABP=\mathrm{\angle }CBQ$ $\therefore$ Vertically Opposite angles

Therefore, by AAS congruence criteria of a triangle, we have

$\mathrm{\Delta }ADE\cong \mathrm{\Delta }CFE$.

By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), 

$PA=QC$ and $AB=BC$ …(1)

Now, we have

$PA=AR$ A is the midpoint of PR. …(2)

$PA=QC$ from (1) …(3)

We can say that 

$PA=AR=CQ$   …from (1),(2),(3)

As AC || RQ, AR || CQ, and AR = CQ, we can say that ACQR is a parallelogram. 

Therefore, AC = RQ. 

Here, $AB=\frac{1}{2}$ AC and AC = RQ

Hence, $AB=\frac{1}{2}$ RQ

Also, as ACRQ, we have ABRQ 

Hence proved.

Converse of Midsegment Theorem of a Triangle

In $\mathrm{\Delta }PQR$, if A is the midpoint of sides PR, AB || RQ and  $AB=\frac{1}{2}RQ$, then AB is the midsegment of the $\mathrm{\Delta }PQR$.

Proof of Converse of Midsegment Theorem

Given: In triangle PQR, AB joins two sides PR and PQ of a triangle PQR.

AB || QR

$AB=\frac{1}{2}\times QR$

To prove: AB is the midsegment of $\mathrm{\Delta }PQR$.

OR 

B is the midpoint of PQ and A is the midpoint of PR.

Construction: Extend the line AB from the point B to the point C, such that $AB=BC$.

Construct a line QC parallel to AR, such that both lines meet at point C.

It is given that A is the midpoint of PR, AB || RQ and  AB $=\frac{1}{2}$ RQ.

$AB=BC$ …construction

$AB=\frac{1}{2}RQ$ …Given

Thus, $AC=QR$

Also, AR || CQ …Given

         AB || RQ …Construction

Therefore, DBCF is a parallelogram.

We know that opposite sides of a parallelogram are parallel and equal.

Therefore, $AR=CQ$ and PA || CQ …(1)

In $\mathrm{\Delta }PAB$ and $\mathrm{\Delta }QBC$,

$\mathrm{\angle }APB=\mathrm{\angle }CQB$ (Alternate angles) 

$AB=BC$    

$\mathrm{\angle }PBA=\mathrm{\angle }QBC$ (Vertically opposite angle)  

So, by AAS congruence criterion, $\mathrm{\Delta }PAB\cong \mathrm{\Delta }QBC$

Using CPCTC, we write

$PB=BQ$

$CQ=AP$

But $AR=CQ$  …from (1)

Thus, $AP=AR$

Therefore, AB is the midsegment of $\mathrm{\Delta }PQR$.

How to Find the Midsegment of a Triangle

To construct the midsegment of a triangle, we find midpoints of any two sides of the triangle.

Step 1: Put the compass on the vertex B. Keep the compass distance more than half the length of BC. Draw arcs on both sides of BC. Draw similar arcs keeping the compass on the vertex C to intersect the previous arcs at points D and G respectively. Draw a line joining the points D and G. Let the line DG intersect the side BC at the point T. T is the midpoint of the side BC.

Step 2: Similarly, find the midpoint of the side AB. Let S be the midpoint of side AB.

Step 3: Join S and T. The segment joining S and T is the midsegment of the triangle ABC.

Midsegment Formula

The formula to find the length of the midsegment is

Length of the midsegment $=\frac{1}{2}\times$ Length of the base of triangle

Facts about the Midsegment of a Triangle

Conclusion

In this article, we learned about the midsegment of a triangle. It is the line that joins the midpoints or center of any two sides of a triangle. Let’s solve a few examples of the midsegment of a triangle for a better understanding!

Solved Examples on Midsegment of a Triangle

1. Find the value of x if MN is the midsegment of the triangle XYZ. 

Solution:  According to the midsegment theorem, the length of the midsegment is half the length of the third side.

$MN=\frac{1}{2}\times QR$

QR $=2$ MN 

$10x=2\times 35$

$10x=70$

$x=7$ units

2. In the triangle below it is given that BC $=36$cm, and M, N are the midpoints of AB and AC. Find XY.

Solution: According to the midsegment theorem, the length of the midsegment is half the length of the third side.

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

$XY=\frac{1}{2}\times 36$

$XY=18$ in

3. AB is the midsegment of the triangle ABC. Find the value of x.

Solution: 

According to the midsegment theorem, the length of the midsegment is half the length of the third side.

$AB=\frac{1}{2}\times XZ$

$3x-1=\frac{1}{2}\times 34$

$3x-1=17$

$3x=18$

$x=6$ feet

4. If ABC is an equilateral triangle with a midsegment of length 12 units then find the perimeter of the triangle ABC. 

Solution: 

According to the midsegment theorem,

Length of the midsegment $=\frac{1}{2}\times$ length of the base of triangle

Let the side of the equilateral triangle be x.

$12=\frac{1}{2}\times x$

$x=2\times 12$

$x=24$ units

Perimeter of an equilateral triangle $=4x=4\times 24=96$ units

5. Find the perimeter of $\mathrm{\Delta }DEF$, if the perimeter of $\mathrm{\Delta }ABC$ is 72 units. 

DE, EF, and DF are midsegments.

Solution:

By the midsegment theorem, we have

DE || BC and DE $=\frac{1}{2}\times$ BC  —————— (i)

Similarly, 

DF || AC and DF $=\frac{1}{2}\times$ AC  —————— (ii)

and 

EF || AB and EF $=\frac{1}{2}\times$ AB  —————— (iii)

Now, the perimeter of $\mathrm{\Delta }DEF=DE+EF+FD$

$\Rightarrow$ Perimeter of $\mathrm{\Delta }DEF=\frac{1}{2}BC+\frac{1}{2}AC+\frac{1}{2}AB$  (from equation i, ii and iii)

$\Rightarrow$ Perimeter of $\mathrm{\Delta }DEF=\frac{1}{2}(BC+AC+AB)$

$\Rightarrow$ Perimeter of $\mathrm{\Delta }DEF=\frac{1}{2}$ $(\text{Perimeter of}\mathrm{\Delta }ABC)=\frac{1}{2}\times 72=36$ cm.

Practice Problems on Midsegment of a Triangle

Frequently Asked Questions about Midsegment of a Triangle