The coordinates of the mid-point of a line segment are the average of the coordinates of its end points.
Midpoint theorem

Consider a line segment $\overline{AB}$ with end points A($x_1$, $y_1$ ) and B($x_2$, $y_2$). Let M(x, y) be a point on the line segment $\overline{AB}$ which divides AB in 1:$\lambda$ ratio where $\lambda$ being a non-zero real number. Let C, D, and E be the foot of perpendiculars of A, B, and M, respectively. Join AC, ME, and BD. Let F be the foot of the perpendicular of A on ME, and G be the foot of the perpendicular of M on BD.
Midpoint Theorem Picture
From the figure, it is clear that
AF = $x - x_1$,
MF = $y - y_1$,
MG = $x_2 - x$,
BG = $y_2 - y$
Moreover, $\Delta$AFM and $\Delta$MGB are similar triangles. Thus, we have

$\frac{AM}{MB}$ = $\frac{AF}{MG}$ = $\frac{MF}{BG}$

$\rightarrow$ $\frac{1}{\lambda}$ = $\frac{(x - x_1)}{(x_2 - x)}$ = $\frac{(y - y_1)}{(y_2-y)}$

$\rightarrow$ ($x_2-x$) = $\lambda$($x - x_1$ ) and ($y_2 - y$) = $\lambda$($y - y_1$)

$\rightarrow$ $x_2 - x$ = $\lambda x - \lambda x_1$ and $y_2 - y$ = $\lambda y - \lambda y_1$

$\rightarrow$ (1+ $\lambda$) x = $\lambda x_1$ + $x_2$ and (1 + $\lambda$) y = $\lambda y_1$ + $y_2$

$\rightarrow$ x = $\frac{\lambda x_1 + x_2}{1 + \lambda}$ and y = $\frac{\lambda y_1 + y_2}{1+ \lambda}$

Thus, we have
M = (x, y) = ($\frac{\lambda x_1 + x_2}{m + n}$, $\frac{\lambda y_1 + y_2}{m + n}$)

It is to be noted that if M is the mid-point of the line segment $\overline{AB}$ , it bisects $\overline{AB}$ in 1:1 ratio. So, we have $\lambda$=1. Substituting in the above formula, we get
Mid-point of $\overline{AB}$ = ($\frac{x_1 + x_2}{2}$, $\frac{y_1 + y_2}{2}$)Hence, we proved the theorem.

Solved Example

Question: Let A (-2, 3) and B(6, 7) be the mid points of a line segment $\overline{AB}$. Find the mid-point of $\overline{AB}$.
Given that
        A($x_1, y_1$) = (-2, 3) and B($x_2, y_2$) = (6, 7).
Then, the mid-point of $\overline{AB}$ is

          M=($\frac{x_1 + x_2}{2}$, $\frac{y_1 + y_2}{2}$)

               =($\frac{-2 + 6}{2}$, $\frac{3 + 7}{2}$)

               =($\frac{4}{2}$, $\frac{10}{2}$)

               =(2, 5)

Statement: The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and its length is equal to the half of the length of the third side.

Triangle Midpoint Theorem

Construction: Let $\Delta$ABC be the given triangle. Let D and E be the mid-points of AB and AC, respectively. Let us join $\overline{DE}$.

To Prove: $\overline{BC}$ || $\overline{DE}$ and DE = $\frac{1}{2}$ (BC).


1.$\frac{AD}{AB}$ = $\frac{AE}{AC}$ = $\frac{1}{2}$D and E are mid points of $\overline{AB}$ and $\overline{AC}$, respectively
2.∠BAC ≅ ∠DAE Common angle
3.ΔABC ~ ΔADE are similar trianglesSAS property of similarity
4.∠ABC ≅ ∠ADE and ∠ACB ≅ ∠AEDCorresponding angles of similar triangles
5.$\overline{BE}$∥$\overline{DE}$Corresponding angles
6.$\frac{DE}{BE}$ = $\frac{AD}{AB}$ = $\frac{1}{2}$From (1) & corresponding sides of similar triangles.
7DE = $\frac{1}{2}$(BE)From (6)

Converse of the Triangle Mid-point Theorem
The line through the mid-point of one side of a triangle when drawn parallel to a second side bisects the third side.

This is a special case of more generalized Triangle Intercept Theorem, which is proved hereunder:

Triangle Intercept Theorem
Statement: A line parallel to one side of a triangle divides the other two sides proportionally.

Construction: Consider a triangle $\Delta$ABC. Draw a line L parallel to $\overline{BC}$ such that it intersects the remaining two sides $\overline{AB}$ and $\overline{AC}$ at D and E, respectively.

To Prove: $\frac{AD}{AB}$ = $\frac{AE}{AC}$

1$\overline{BC}$ ∥ $\overline{DE}$Given
2∠ADE ≅ ∠ABCCorresponding angles
3∠DAE ≅ ∠BACCommon angle
4ΔADE ~ ΔABCAA Property of similarity
5$\frac{AD}{AB}$ = $\frac{AE}{AC}$Corresponding sides of Similar Triangles

In case of the converse of the triangle mid-point theorem, if we take D to be mid-point of AB, then we have
$\frac{AD}{AE}$ = $\frac{AE}{AC}$

$\rightarrow$ $\frac{1}{2}$ = $\frac{AE}{AC}$

$\rightarrow$ AE = $\frac{1}{2}$ AC

That is, E is the mid-point of AC.

Solved Example

Question: Use the triangle mid-point theorem to find out the value of x from the following figure.

Triangle Midpoint Theorem Example
From the Triangle Mid-point Theorem, we have

DE = $\frac{1}{2}$ BC

$\rightarrow$ BC = 2 DE

$\rightarrow$ x = 2(2.5)

$\rightarrow$ x = 5 cm