Question

State and prove Basic Proportionality theorem.

Answer 1

Basic Proportionality Theorem states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points,then the line divides those sides of the triangle in proportion.

Let $ABC$ be the triangle.

The line $l$ parallel to $BC$ intersect $AB$ at $D$ and $AC$ at $E$.

To prove $\frac{AD}{DB}=\frac{AE}{EC}$

Join $BE,CD$

Draw $EF\perp AB$, $DG\perp CA$

Since $EF\perp AB$,

$EF$ is the height of triangles $ADE$ and $DBE$

Area of $△ADE=\frac{1}{2}\times$ base $\times$ height$=\frac{1}{2}AD\times EF$

Area of $△DBE=\frac{1}{2}\times DB\times EF$
$\frac{areaof\Delta ADE}{areaof\Delta DBE}=\frac{\frac{1}{2}\times AD\times EF}{\frac{1}{2}\times DB\times EF}=\frac{AD}{DB}$ ........(1)
Similarly,
$\frac{areaof\Delta ADE}{areaof\Delta DCE}=\frac{\frac{1}{2}\times AE\times DG}{\frac{1}{2}\times EC\times DG}=\frac{AE}{EC}$ ......(2)

But $\Delta DBE$ and $\Delta DCE$ are the same base $DE$ and between the same parallel straight line $BC$ and $DE$.

Area of $\Delta DBE=$ area of $\Delta DCE$ ....(3)

From (1), (2) and (3), we have

$\frac{AD}{DB}=\frac{AE}{EC}$

Hence proved.