Class 9 Mid-Point Theorem

Statement:

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

Given:

In $\Delta \text{ABC}$, 

$\hspace{1 cm}\text{AD} = \text{DB}\hspace{1.5 cm}(1)$

$\hspace{1 cm}\text{AE} = \text{EC}\hspace{1.6 cm}(2)$

To Prove:

$\text{DE} ∥ \text{BC}$ and $\text{DE}=\frac{1}{2}\text{ BC}$

Construction:

Extend $\text{DE}$ to $\text{F}$ such that $\text{BA} ∥ \text{CF}$.

Proof:

In $\Delta \text{DAE}$ and $\Delta \text{FCE}$

$\hspace{1.5 cm}\angle 1 = \angle 4$

(If lines are parallel then Alternate Interior Angles are equal.)

$\hspace{1.5 cm}\text{AE} = \text{CE} \hspace{1 cm}$(Given)

$\hspace{1.6 cm}\angle 2 = \angle 3$ 

(Vertically Opposite Angles.)

$\text{So,}\;\Delta\text{DAE}\cong\Delta\text{FCE}\hspace{0.3 cm}(\text{By ASA Rule})$

By CPCTC,

$\hspace{1.5 cm}\text{AD}=\text{CF}\hspace{1.5 cm}(3)$

$\hspace{1.5 cm}\text{DE}=\text{FE}\hspace{1.6 cm}(4)$

From $(1)$ and $(3)$, we get

$\hspace{1.5 cm}\text{BD}=\text{CF}$

$\hspace{1.5 cm}\text{BD}\; ||\; \text{CF}\hspace{1.6 cm}(\text{BA}\;||\;\text{CF})$

So, $\text{BDFC}$ is a parallelogram.

(In a quadrilateral, if one pair of opposite sides are equal and parallel then it becomes a parallelogram.)

$\implies\hspace{1.5 cm} \text{DF} \;||\; \text{BC}$

(Opposite sides of Parallelogram are parallel.)

$\hspace{2.6 cm}\boxed{\text{DE}\;||\;\text{BC}}\hspace{1.5 cm}(5)$

$\implies\hspace{1.5 cm} \text{DF} = \text{BC}$

(Opposite sides of Parallelogram are equal.)

$\hspace{1.5 cm}\text{DE}+\text{EF}=\text{BC}$

$\hspace{1.5 cm}\text{DE}+\text{DE}=\text{BC}\hspace{0.75 cm}[\text{From }(4)]$

$\hspace{2 cm}2\times\text{DE}=\text{BC}$

$ \hspace{2.7 cm}\boxed{\text{DE}=\frac{1}{2}\text{BC}}\hspace{1 cm}(6)$

$\hspace{2.4 cm}$ Hence Proved.