Triangle has numerous subparts. Such as, angles, sides, median, midpoint, midsegment, etc. Here's an task for you. You deserve to now visualize various types of triangles in math based upon their sides and angles. Try changing the place of the vertices to understand the relationship between sides and angles the a triangle.

You are watching: Definition of midsegment of a triangle

In the later part of this chapter we will certainly discuss around midpoint and midsegments of a triangle.

Given any type of two points, speak \(A\) and also \(C\), the midpoint is a allude \(B\) i beg your pardon is situated halfway between the points\(A\) and \(B\).


Observe that the point\(B\)is equidistant from\(A\) and also \(C\).

A midpoint exists only for a line segment.

Line which connects the midpoint is termed together midsegment.

In this mini-lesson, we will check out the civilization of midsegment the a triangle by recognize the answers to the questions prefer what is midsegment of a triangle, triangle midsegment theorem, and proof through the assist of interaction questions.

So let's acquire started!

Lesson Plan

1.What Is Midsegment the a Triangle?
2.Important notes on Midsegment of a Triangle
3.Solved instances on Midsegment of a Triangle
4.Interactive concerns onMidsegment the a Triangle
5.Challenging inquiry on Midsegment of a Triangle

What Is Midsegment the a Triangle?

Midsegment the a TriangleDefinition

A midsegment the a triangle is a line segment the joinsthe midpoints or center of 2 opposite or surrounding sides the a triangle


In the over figure, D is the midpoint the ABand E is the midpoint that AC.

Here DE is a midsegment the a triangle ABC.

Triangle Midsegment Theorem

The midsegment theorem states that aline segmentconnectingthe midpoints of anytwo sides of a triangle is parallel come 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 also \(DE=\dfrac12\ BC\)


\(AD=DB\)  and  \(BF=FC\)

Then according to the midsegmenttheorem

\(DF∥AC\) and \(DF=\dfrac12\ AC\)


\(AE=EC\)  and  \(BF=FC\)

Then follow to the midsegmenttheorem

\(EF∥AB\) and \(EF=\dfrac12\ AB\)

Proof for Midsegment the a Triangle

In the over section, we observed a triangle \(ABC\), v \(D,\) \(E,\) and \(F\) as 3 midpoints.

We should prove two things to justification the proof ofthe triangle midsegment theorem:

\(DE∥BC\)\(DE=\dfrac12\ BC\)


Given:D and also E space the midpoints of abdominal muscle and AC

To prove,\(DE∥BC\) and also \(DE=\dfrac12\ BC\) we need to draw a heat parallel to ab meet E produced at F.

In \(\bigtriangleupADE\) and also \(\bigtriangleupCFE\)

\(\beginalign AE &=EC\text (E is the midpoint of AC)\\\ \angle1 &=\angle2\text (Vertically the opposite angles)\\\ \angle3 &=\angle4\text (Alternate angles)\endalign\)

By AAS congruency that triangle us have,

\(\bigtriangleupADE \cong \bigtriangleupCFE\)

By CPCT us have,



D is the midpoint of AB



DBCF is a parallelogram,

\(DF || BC\) and \(DF = BC\)

\(DE || BC\) and also \(DF = BC\)


since, DF = BC


Hence Proved

Midsegment of a Triangle Formula

Midsegment \(=\) \(\dfrac12\times\) Triangle Base

What Is the Converse of the Triangle Midsegment Theorem?

The converse the the midsegment to organize is identified as: Whena line segmentconnects twomidpoints of two opposite political parties of a triangle and is parallel to the 3rd side of a triangleand is fifty percent of it climate it is a midsegment that a triangle.


Intriangle alphabet we have,

\(DE∥BC\) and \(DE=\dfrac12\ BC\)

Then follow to the converse that thetriangle midsegmenttheorem

\(AD=DB\) and also \(AE=EC\)\(DE\) is a midsegment the triangle \(ABC\)

Proof for Converse of the TriangleMidsegment Theorem

In the above section, we witnessed \(\bigtriangleupABC\), v \(D,\) \(E,\) and \(F\) as three midpoints.

We need to prove any type of one ofthe things mentioned below to justify the proof ofthe converse the a triangle midsegment theorem:

\(DE\) is a midsegment of a \(\bigtriangleupABC\)\(AD=DB\) and also \(AE=EC\)


We have actually D together the midpoint of AB, then\(AD = DB\) and \(DE||BC\)

\(AB\) \(=\) \(AD + DB\) \(=\) \(DB + DB\) \(=\) \(2DB\)

DBCF is a parallelogram.

\(DE||BC\) and also \(BD||CF\)

Opposite sides of a parallelogram space equal.



In \(\bigtriangleupADE\) and also \(\bigtriangleupCFE\)

\(\beginalign\angle1 &=\angle2\text (Vertically opposite angles)\\\ \angle3 &=\angle4\text (Alternate angles)\\\ DA &=CF\endalign\)

By AAS congruency that triangle we have,

\(\bigtriangleupADE \cong \bigtriangleupCFE\)

By CPCT we have


E is the midpoint of AC and DF.

Hence, DE is a midsegment that \(\bigtriangleupABC\).


Example 1

In the given figure H and M room 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 = 3b)EH = 16, FH = 12, EM = 4and GM = 3



a) us haveEH = 6, FH = 9, EM = 2, and also GM = 3


\(\dfracEMGM= \dfracEHFH=\dfrac23\)

b)We haveEH = 16, FH = 12, EM = 4,and GM = 3


\(\dfracEMGM= \dfracEHFH=\dfrac43\)

HM divides EF and also EG that triangle EFG in equal ratios.

Hence, HM is themidsegment of triangle EFG.

\(\therefore\) \(HM || FG\)

Example 2

Help Ron in detect the value of xand the value of line segmentAB, given that A and B space midpoints that triangle PQR.



We have two midpoints A and B.

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According to the midsegment triangle theorem

\(\beginalignQR &=2AB\\\36 &=2(9x)\\\x &=2\\\AB &=18\endalign\)

\(\therefore\) The worth of x is 2The worth of abdominal muscle is 18

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