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.

(N/A) Let a $\triangle ABC$ in which a line $DE$ parallel to $BC$ intersects $AB$ at $D$ and $AC$ at $E$. To prove: $DE$ divides the two sides in the same ratio,i.e.,$\frac{AD}{DB} = \frac{AE}{EC}$.
Construction: Join $BE, CD$ and draw $EF \perp AB$ and $DG \perp AC$.
Proof: Here,$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EF}{\frac{1}{2} \times DB \times EF} = \frac{AD}{DB}$ ......$(i)$ [Since,area of triangle $= \frac{1}{2} \times \text{base} \times \text{height}$]
Similarly,$\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DG}{\frac{1}{2} \times EC \times DG} = \frac{AE}{EC}$ ......$(ii)$
Now,since $\triangle BDE$ and $\triangle DEC$ lie between the same parallel lines $DE$ and $BC$ and on the same base $DE$,therefore,$\text{ar}(\triangle BDE) = \text{ar}(\triangle DEC)$ ......$(iii)$
From Eqs. $(i), (ii)$ and $(iii)$,we get $\frac{AD}{DB} = \frac{AE}{EC}$.
Hence proved.