Question

STATE BPT THEOREM AND PROVE IT.

Answer 1

Basic proportionality theorem :

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

Given:

A $△$ ABC in which $DE\parallel BC$ and DE intersects AB and AC at D and E respectively.

To prove :

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

Construction :

Join BE and CD.

Draw $EL\perp AB$ and $DM\perp AC$.

Proof :

We have,

$ar\left(△ADE\right)=\frac{1}{2}\times AD\times EL$

$ar\left(△DBE\right)=\frac{1}{2}\times DB\times EL$

Therefore,

$\frac{ar\left(△ADE\right)}{ar\left(△DBE\right)}=\frac{AD}{DB}$ ...(1)

Again,

$ar\left(△ADE\right)=ar\left(△AED\right)=\frac{1}{2}\times AE\times DM$

$ar\left(△ECD\right)=\frac{1}{2}\times EC\times DM$

Therefore,

$\frac{ar\left(△ADE\right)}{ar\left(△ECD\right)}=\frac{AE}{EC}$ ....(2)

Now, $△DBE$ and $△ECD$ being on the same base DE and between the same parallels DE and BC, we have,

$ar\left(△DBE\right)=ar\left(△ECD\right)$ ....(3)

From 1, 2 & 3, we have,

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