A closed figure made with 3 line segments forms the shape of a triangle.
Let us explore the world of triangle
Triangle has many subparts. Such as, angles, sides, median, midpoint, midsegment, etc. Here's an activity for you. You can now visualize various types of triangles in math based on their sides and angles. Try changing the position of the vertices to understand the relationship between sides and angles of a triangle.
In the later part of this chapter we will discuss about midpoint and midsegments of a triangle.
Given any two points, say \(A\) and \(C\), the midpoint is a point \(B\) which is located halfway between the points\(A\) and \(B\).
Observe that the point\(B\)is equidistant from\(A\) and \(C\).
A midpoint exists only for a line segment.
Line which connects the midpoint is termed as midsegment.
In this mini-lesson, we will explore the world of midsegment of a triangle by finding the answers to the questions like what is midsegment of a triangle, triangle midsegment theorem, and proof with the help of interactive questions.
So let's get started!
Lesson Plan
| 1. | What Is Midsegment of a Triangle? |
| 2. | Important Notes on Midsegment of a Triangle |
| 3. | Solved Examples on Midsegment of a Triangle |
| 4. | Interactive Questions onMidsegment of a Triangle |
| 5. | Challenging Question on Midsegment of a Triangle |
What Is Midsegment of a Triangle?
Midsegment of a TriangleDefinition
A midsegment of a triangle is a line segment that joinsthe midpoints or center of two opposite or adjacent sides of a triangle
In the above figure, D is the midpoint of ABand E is the midpoint of AC.
Here DE is a midsegment of a triangle ABC.
Triangle Midsegment Theorem
The midsegment theorem states that aline segmentconnectingthe midpoints of anytwo sides of a triangle is parallel to the third side of a triangleand is half of it.
In the triangle ABC we have,
\(AD=DB\) and \(AE=EC\)
Then according to the midsegmenttheorem
\(DE∥BC\) and \(DE=\dfrac{1}{2}\ BC\)
Similarly,
\(AD=DB\) and \(BF=FC\)
Then according to the midsegmenttheorem
\(DF∥AC\) and \(DF=\dfrac{1}{2}\ AC\)
Similarly,
\(AE=EC\) and \(BF=FC\)
Then according to the midsegmenttheorem
\(EF∥AB\) and \(EF=\dfrac{1}{2}\ AB\)
Proof for Midsegment of a Triangle
In the above section, we saw a triangle \(ABC\), with \(D,\) \(E,\) and \(F\) as three midpoints.
We need to prove two things to justify the proof ofthe triangle midsegment theorem:
- \(DE∥BC\)
- \(DE=\dfrac{1}{2}\ BC\)
Given:D and E are the midpoints of AB and AC
To prove,\(DE∥BC\) and \(DE=\dfrac{1}{2}\ BC\) we need to draw a line parallel to AB meet E produced at F.
In \(\bigtriangleup{ADE}\) and \(\bigtriangleup{CFE}\)
\(\begin{align} AE &=EC\text{ (E is the midpoint of AC)}\\\ \angle{1} &=\angle{2}\text{ (Vertically opposite angles)}\\\ \angle{3} &=\angle{4}\text{ (Alternate angles)}\end{align}\)
By AAS congruency of triangle we have,
\(\bigtriangleup{ADE} \cong \bigtriangleup{CFE}\)
By CPCT we have,
\(DE=FE\)
\(AD=CF\)
D is the midpoint of AB
\(AD=BD\)
\(BD=CF\)
DBCF is a parallelogram,
\(DF || BC\) and \(DF = BC\)
\(DE || BC\) and \(DF = BC\)
\(DE=\dfrac{1}{2}DF\)
since, DF = BC
\(DE=\dfrac{1}{2}BC\)
Hence Proved
Midsegment of a Triangle Formula
Midsegment \(=\) \(\dfrac{1}{2}\times\) Triangle Base
What Is the Converse of the Triangle Midsegment Theorem?
The converse of the midsegment theorem is defined as: Whena line segmentconnects twomidpoints of two opposite sides of a triangle and is parallel to the third side of a triangleand is half of it then it is a midsegment of a triangle.
Intriangle ABC we have,
\(DE∥BC\) and \(DE=\dfrac{1}{2}\ BC\)
Then according to the converse of thetriangle midsegmenttheorem
\(AD=DB\) and \(AE=EC\)
\(DE\) is a midsegment of triangle \(ABC\)
Proof for Converse of the TriangleMidsegment Theorem
In the above section, we saw \(\bigtriangleup{ABC}\), with \(D,\) \(E,\) and \(F\) as three midpoints.
We need to prove any one ofthe things mentioned below to justify the proof ofthe converse of a triangle midsegment theorem:
- \(DE\) is a midsegment of a \(\bigtriangleup{ABC}\)
- \(AD=DB\) and \(AE=EC\)
We have D as the midpoint of AB, then\(AD = DB\) and \(DE||BC\)
\(AB\) \(=\) \(AD + DB\) \(=\) \(DB + DB\) \(=\) \(2DB\)
DBCF is a parallelogram.
\(DE||BC\) and \(BD||CF\)
Opposite sides of a parallelogram are equal.
\(BD=CF\)
\(DA=CF\)
In \(\bigtriangleup{ADE}\) and \(\bigtriangleup{CFE}\)
\(\begin{align}\angle{1} &=\angle{2}\text{ (Vertically opposite angles)}\\\ \angle{3} &=\angle{4}\text{ (Alternate angles)}\\\ DA &=CF\end{align}\)
By AAS congruency of triangle we have,
\(\bigtriangleup{ADE} \cong \bigtriangleup{CFE}\)
By CPCT we have
\(AE=EC\)
E is the midpoint of AC and DF.
Hence, DE is a midsegment of \(\bigtriangleup{ABC}\).
Important Notes
a)The line segment through a midpointis always parallel to oneside of the triangle.
b) The midsegment \(=\) \(\dfrac{1}{2}\) the length of the third side of a triangle.
c) A triangle can have a maximum of threemidsegments.
d) The midsegment of a triangle theorem is also known as mid-point theorem.
Solved Examples
To understand the midsegment of a triangle better,let us look at some solved examples.
Example 1
In the given figure H and M are the midpoints of triangle EFG. Help Jamie to prove \(HM||FG\) for the following two cases.
a) EH = 6, FH = 9, EM = 2 and GM = 3
b)EH = 16, FH = 12, EM = 4and GM = 3
Solution
a) We haveEH = 6, FH = 9, EM = 2, and GM = 3
\(\dfrac{EH}{FH}=\dfrac{6}{9}=\dfrac{2}{3}\)
\(\dfrac{EM}{GM}= \dfrac{EH}{FH}=\dfrac{2}{3}\)
b)We haveEH = 16, FH = 12, EM = 4,and GM = 3
\(\dfrac{EH}{FH}=\dfrac{16}{12}=\dfrac{4}{3}\)
\(\dfrac{EM}{GM}= \dfrac{EH}{FH}=\dfrac{4}{3}\)
HM divides EF and EG of triangle EFG in equal ratios.
Hence, HM is themidsegment of triangle EFG.
\(\therefore\) \(HM || FG\)
Example 2
Help Ron in finding the value of xand the value of line segmentAB, given that A and B are midpoints of triangle PQR.
Solution
We have two midpoints A and B.
According to the midsegment triangle theorem
\(\begin{align}QR &=2AB\\\
36 &=2(9x)\\\
x &=2\\\
AB &=18\end{align}\)
\(\therefore\) The value of x is 2
The value of AB is 18
Interactive Questions
Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.
Challenging Questions
a)Consider a triangle ABC, and let D be any point on BC. Let X and Y be the midpoints of AB and AC.Show that XY will bisect AD.
b)Consider a parallelogram ABCD. E and F are the midpoints of AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.
Let's Summarize
The mini-lesson targetedthe fascinating concept of the midsegment of a triangle. The math journey aroundthe midsegment of a trianglestarts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.
About Cuemath
AtCuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!
Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.
Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.
Frequently Asked Questions(FAQ's)
1. What is the midsegment of a triangle?
A midsegment of a triangle is a line segment that joinsthe midpoints or center of two opposite or adjacent sides of a triangle
In atriangle, we can have 3 midsegments.
In the above figure, D is the midpoint of ABand E is the midpoint of AC, and F is the midpoint of BC.
Here DE, DF, and EF are 3 midsegments of a triangle ABC.
2. How to find the midsegment of a triangle?
We can find the midsegment of a triangle by using the midsegment of a triangle formula,
Midsegment \(=\) \(\dfrac{1}{2}\times\) Triangle Base.
3. What is the midpoint theorem?
The midpoint theorem statesthatthe line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of the third side. Consider an arbitrary triangle, \(\bigtriangleup{ABC}\). Let D and E be the midpoints of AB and AC. Suppose that you join D and E:
The midpoint theorem says that DE will be parallel to BC and equal to exactly half of BC.
