Question

State $B.P.T$ and prove it.

Answer 1

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

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

Construction:Join $BE$ and $CD$.Draw $DM\perp AC$ and $EN\perp AB$

Proof:

Now, ar$\left(△ADE\right)=\frac{1}{2}\times AD\times EN$ ......$\left(1\right)$

ar$\left(△BDE\right)=\frac{1}{2}\times DB\times EN$ ......$\left(2\right)$

ar$\left(△ADE\right)=\frac{1}{2}\times AE\times DM$ ......$\left(3\right)$

ar$\left(△DEC\right)=\frac{1}{2}\times EC\times DM$ ......$\left(4\right)$

Divide $\left(1\right)$ and $\left(2\right)$

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

$\frac{ar\left(△ADE\right)}{ar\left(△BDE\right)}=\frac{AD}{DB}$ ..$\left(5\right)$

Divide $\left(3\right)$ and $\left(4\right)$

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

$\frac{ar\left(△ADE\right)}{ar\left(△DEC\right)}=\frac{AE}{EC}$ ..$\left(6\right)$

Now, $△BDE$ and $△DEC$ are on the same base $DE$ and between the same parallel lines $BC$ and $DE$

$\therefore ar\left(△BDE\right)=ar\left(△DEC\right)$

Hence,$\frac{ar\left(△ADE\right)}{ar\left(△DEC\right)}=\frac{ar\left(△ADE\right)}{ar\left(△BDE\right)}$

$\Rightarrow \frac{AD}{DB}=\frac{AE}{EC}$ from $\left(5\right)$ and $\left(6\right)$

Hence proved.