This topic gives an overview of 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 :
Repeat this activity with some more triangles. So, you arrive at the following theorem
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
You will see that converse of the above theorem is also true which is stated as below:
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.
Cite this Simulator: