Midpoint Theorem

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

Prerequisite:
Parallelogram (definition)
Vertical Angle Theorem (proof)
Alternate angles property (proof)
SAS congruency (proof)
Characteristics of parallelogram (proof)

Proof:

Let ABC be a triangle. Let DE be a line such that D and E are the midpoints of the sides AB and AC respectively. To prove that DE is parallel to BC and equal to half of its length, let us extend the line DE to F, such that DE $=$ EF. Join C and F by a straight line.

Now, in $\triangle ADE$ and $\triangle CFE$,

$\qquad\quad AE = CE\qquad\qquad\qquad\qquad\qquad\text{(given)}$
$\qquad\quad DE = EF\qquad\qquad\qquad\qquad\qquad\text{(by construction)}$
$\qquad\quad\angle AED = \angle CEF\qquad\quad\qquad\qquad\text{(vertically opposite angles)}$

Hence, $\triangle ADE\cong\triangle CFE$ by SAS rule.

Therefore by CPCTC, $\angle DAE = \angle FCE$, which are the alternate angles. Hence, by alternate angles property, AB $\parallel$ CF. Further, AD $=$ CF.

Now, in quadrilateral DBCF,

$\qquad\quad BD = CF (= AD)\qquad\qquad\qquad\quad\text{(since D is the midpoint of AB)}\\ \qquad\quad BD \parallel CF\qquad\qquad\qquad\qquad\qquad\;\:\:\text{(proved above)}$

Hence, DBCF is a parallelogram (see characteristics of parallelogram).

Therefore, by definition of a parallelogram, DF $\parallel$ BC and hence DE $\parallel$ BC. Also, by properties (characteristics) of a parallelogram, DF $=$ BC.

Since, by construction, DE $=$ DF, hence,

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

Q.E.D.


Recommended:
Converse of Midpoint Theorem
Basic Proportionality Theorem
Pythagoras Theorem