Basic Proportionality Theorem

Statement: 

If a line is drawn parallel to the one side of the triangle to intersect other two sides of the triangle in distinct points, then other two sides are divided in the same ratio.

Given: 

$\mathrm{\Delta }\text{ABC},\text{PQ}||\text{BC}$

To Prove: 

$\frac{\text{AP}}{\text{BP}}=\frac{\text{AQ}}{\text{CQ}}$

Construction: 

$\text{QM}\perp \text{AB},\text{PN}\perp \text{AC},$ Join $\text{P}$ & $\text{C}$ and $\text{B}$ & $\text{Q}$.

Proof: 

Two triangles on the same base and between the same parallels are equal in area.

So, $ar(\mathrm{\Delta }\text{BPQ})=ar(\mathrm{\Delta }\text{CPQ})$          (1)

Area of triangle $=\frac{1}{2}\times$ Base $\times$ Height

$ar(\mathrm{\Delta }\text{APQ})=\frac{1}{2}\times \text{AP}\times \text{MQ}$        (2)

$ar(\mathrm{\Delta }\text{BPQ})=\frac{1}{2}\times \text{BP}\times \text{MQ}$        (3)

$ar(\mathrm{\Delta }\text{APQ})=\frac{1}{2}\times \text{AQ}\times \text{PN}$        (4)

$ar(\mathrm{\Delta }\text{CPQ})=\frac{1}{2}\times \text{CQ}\times \text{PN}$         (5)

From (1), (2), (3), (4) and (5), we get

$\frac{ar(\mathrm{\Delta }\text{APQ})}{ar(\mathrm{\Delta }\text{BPQ})}=\frac{ar(\mathrm{\Delta }\text{APQ})}{ar(\mathrm{\Delta }\text{CPQ})}$

$\frac{\frac{1}{2}\times \text{AP}\times \text{MQ}}{\frac{1}{2}\times \text{BP}\times \text{MQ}}=\frac{\frac{1}{2}\times \text{AQ}\times \text{PN}}{\frac{1}{2}\times \text{CQ}\times \text{PN}}$

$\frac{\text{AP}}{\text{BP}}=\frac{\text{AQ}}{\text{CQ}}$ 

Hence Proved.