Mid-Point Theorem


This topic gives an overview of Mid-point Theorem.

The Mid-point Theorem

You have studied many properties of a triangle as well as a quadrilateral. Now let us study yet another result which is related to the mid-point of sides of a triangle. Perform the following activity.

Draw a triangle and mark the mid-points Eand F of two sides of the triangle. Join the points E and F.

Measure EF and BC. Measure ∠ AEF and ∠ ABC.

You will find that :

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»E«/mi»«mi»F«/mi»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»B«/mi»«mi»C«/mi»«mo»§nbsp;«/mo»«mi»a«/mi»«mi»n«/mi»«mi»d«/mi»«mo»§nbsp;«/mo»«mo»§#8736;«/mo»«mi»A«/mi»«mi»E«/mi»«mi»F«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§#8736;«/mo»«mi»A«/mi»«mi»B«/mi»«mi»C«/mi»«/math» so, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»E«/mi»«mi»F«/mi»«mo»§#8741;«/mo»«mi»B«/mi»«mi»C«/mi»«/math»

Repeat this activity with some more triangles. So, you arrive at the following theorem

Theorem 1

The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

You can prove this theorem using the following clue:

Observe the figure in which E and F are mid-points of AB and AC respectively and CD || BA.

∆ AEF ≅ ∆ CDF (ASA Rule)

So, EF = DF and BE = AE = DC. Therefore, BCDE is a parallelogram. This gives EF || BC.

In this case, also note that  «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»E«/mi»«mi»F«/mi»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»§nbsp;«/mo»«mi»E«/mi»«mi»D«/mi»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»§nbsp;«/mo»«mi»B«/mi»«mi»C«/mi»«mo».«/mo»«/math»

You will see that converse of the above theorem is also true which is stated as below:

Theorem 2

The line drawn through the mid-point of one side of a triangle,parallel to another side bisects the third side.

Observe that that E is the mid-point of AB, line l is passsing through E and is parallel to BC and CM || BA.

Prove that AF = CF by using the congruence of ∆ AEF and ∆CDF.


In ∆ ABC, D, E and F are respectively the mid-points of sides AB, BC and CA. Show that ∆ ABC is divided into four congruent triangles by joining D, E and F.


As D and E are mid-points of sides AB and BC of the triangle ABC, by Theorem 1,

DE || AC

Similarly, DF || BC and EF || AB

Therefore ADEF, BDFE and DFCE are all parallelograms.

Now DE is a diagonal of the parallelogram BDFE,

 therefore, ∆ BDE ≅ ∆ FED

Similarly ∆ DAF ≅ ∆ FED

and ∆ EFC ≅ ∆ FED

So, all the four triangles are congruent.


l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p. Show that l, m and n cut off equal intercepts DE and EF on q also.


We are given that AB = BC and have to prove that DE = EF.

Let us join A to F intersecting m at G. The trapezium ACFD is divided into two triangles; namely  Δ  ACF and  Δ  AFD. In Δ ACF, it is given that B is the mid-point of AC (AB = BC) and  BG || CF (since m || n). So, G is the mid-point of AF (by using Theorem 2)

Now, in Δ AFD, we can apply the same argument as G is the mid-point of AF, GE || AD and so by Theorem 8.10, E is the mid-point of DF, ie,   DE=EF.  In other words, l, m and  n  cut off equal intercepts on  q  also.


  • The line-segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it.
  • A line through the mid-point of a side of a triangle parallel to another side bisects the third side.
  • The quadrilateral formed by joining the mid-points of the sides of a quadrilateral, in order, is a parallelogram


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