Midpoint Theorem

The midpoint is the point that divides a line segment into two equal parts. In geometry, the midpoint theorem is a theorem that tells us what happens when the midpoints of two sides of a triangle are joined by a line segment that is parallel to the third side.

Thus, the midpoint theorem helps us find the relation between the line segment, which joins the midpoints of any two sides of a triangle and the third. It can also be applied to establish theorems and properties related to polygons like parallelograms, trapezoids, and others.

Statement

The midpoint theorem states that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half the length of the third side.

If we consider △ABC with D and E as the midpoints of AB and AC, respectively, then according to the midpoint theorem

DE || BC and DE = ${\dfrac{1}{2}}$  × BC

Proof

Step 1: A triangle △ABC is drawn where D is the midpoint of AB and E is the midpoint of AC.
Thus, AD = DB ………. (i)  and AE = EC ………. (ii)

Step 2: We draw a line through C parallel to AB such that the extended DE intersects the newly drawn parallel line at point F.

Step 3: In triangles △ADE and △EFC,
AE = EC               [using (ii)]
∠DEA = ∠CEF = ∠1     [Vertically opposite angles]
∠DAE = ∠ECF = ∠2      [Alternate interior angles]
Hence, △ADE ≅ △EFC   [by the AAS congruence criterion]

Step 4: From Step 3,
AD = FC   [Corresponding Parts of Congruent Triangles (CPCT) △ADE and △EFC] ………. (iii)  and
DE = EF [CPCT △ADE and △EFC] ………. (iv)

Hence, DB = FC [using (i) and (iii)] and
DB || FC [by our construction]
Thus ▱ DBCF is a parallelogram.

Step 5:

Using the property of the parallelogram, as DF and BC are the opposite sides of the parallelogram ▱ DBCF,
DF = BC ………. (v) and DF || BC ………. (vi)  
Now, DF = DE + EF [as E lies on the line segment DF] ………. (vii)
This implies that DE ||BC [using (vi) and (vii)] ……….  (viii)
Also DF = DE + EF = 2DE [using (iv) and (vii)] ………. (ix)
Thus, DF = ${\dfrac{1}{2}}$  × BC [using (v) and (ix)] ………. (x)
Hence, the midpoint theorem is proved by (vi) and (x).

Converse of Mid-Point Theorem

According to the converse of the mid-point theorem, if a line drawn through the midpoint of one side of a triangle is parallel to another side, it will bisect the third side.

Let △ABC be a triangle where D is the midpoint of AB. A line through D and parallel to BC intersects AC at E.
So, AD = DB ………. (i) and
DE || BC ………. (ii)

Here, we will prove E is the midpoint of AC, i.e., AE = EC

Proof

Step 1: A line is drawn through C and parallel to AB, which intersects the extended DE at point F. Thus DB || FC ………. (iii)

Step 2: Now, DF || BC [using (ii) and the fact that DF = DE + EF] ………. (iv)
Using (iii) and (iv), we have the parallelogram ▱ DBCF.
Hence, DB = CF [opposite sides of ▱ DBCF] ………. (v) and
DF = BC [opposite sides of ▱ DBCF] ………. (vi)
Using (i) and (v), AD = CF ……….(vii)
Step 3: For △ADE and △EFC,
AD = CF [using (vii)]
∠DAE = ∠ECF     [alternate interior angles]
∠DEA = ∠CEF     [ vertically opposite angles]
Thus,  △ADE ≅ △EFC  [by the AAS congruence criterion] and
AE = EC [corresponding parts of congruent triangles].
Thus, E is the midpoint of AC, which proves the converse of the midpoint theorem.

Formula

The midpoint formula helps to find the midpoint between the two given points.

If M (x₁, y₁) and N (x₂, y₂) are the coordinates of the two given endpoints of a line segment, then the mid-point (x, y) formula will be given by

${\left(\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2}\right)}$

Solved Examples