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 $\mathrm{\Delta }\text{ABC}$, 

$\text{AD}=\text{DB}(1)$

$\text{AE}=\text{EC}(2)$

To Prove:

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

Construction:

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

Proof:

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

$\mathrm{\angle }1=\mathrm{\angle }4$

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

$\text{AE}=\text{CE}$(Given)

$\mathrm{\angle }2=\mathrm{\angle }3$ 

(Vertically Opposite Angles.)

$\text{So,}\mathrm{\Delta }\text{DAE}\cong \mathrm{\Delta }\text{FCE}(\text{By ASA Rule})$

By CPCTC,

$\text{AD}=\text{CF}(3)$

$\text{DE}=\text{FE}(4)$

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

$\text{BD}=\text{CF}$

$\text{BD}||\text{CF}(\text{BA}||\text{CF})$

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

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

$⟹\text{DF}||\text{BC}$

(Opposite sides of Parallelogram are parallel.)

$\boxed{{\displaystyle \text{DE}||\text{BC}}}(5)$

$⟹\text{DF}=\text{BC}$

(Opposite sides of Parallelogram are equal.)

$\text{DE}+\text{EF}=\text{BC}$

$\text{DE}+\text{DE}=\text{BC}[\text{From }(4)]$

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

$\boxed{{\displaystyle \text{DE}=\frac{1}{2}\text{BC}}}(6)$

$$ Hence Proved.