Question

Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.

Answer 1

In $∆ABC$, line $DE$ parallel to $BC$ and intersects $AB$ at $D$ and $AC$ at $E$.

We have to prove that, $DE$ divides the two sides in the same ratio

i.e.,.$\frac{AD}{DB}=\frac{AE}{EC}$

Construction: Join $BE,CD$

Draw $EF\perp AB$ and $DG\perp AC$.

We know that,

$Areaoftriangle=\frac{1}{2}\times base\times height$

$\frac{area\left(∆ADE\right)}{area\left(∆BDE\right)}=\frac{{\displaystyle \frac{1}{2}}\times AD\times EF}{{\displaystyle \frac{1}{2}}\times DB\times EF}$

Therefore, $\frac{area\left(∆ADE\right)}{area\left(∆BDE\right)}=\frac{AD}{DB}$………………$\left(i\right)$

$\frac{area\left(∆ADE\right)}{area\left(∆DEC\right)}=\frac{{\displaystyle \frac{1}{2}}\times AE\times GD}{{\displaystyle \frac{1}{2}}\times EC\times GD}$

Therefore, $\frac{area\left(∆ADE\right)}{area\left(∆DEC\right)}=\frac{AE}{EC}$………………..$\left(ii\right)$

Since, $∆BDE$ and $∆DEC$ lie between the same parallel $DE\mathrm{and}BC$ and are on the same base $DE$.

We obtain, $area(∆BDE)=area(∆DEC)$ …….$\left(iii\right)$

From Equation $\left(i\right),\left(ii\right)\mathrm{and}\left(iii\right)$,

We get, $\frac{AD}{DB}=\frac{AE}{EC}$

Hence, it is proved that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.