Question

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

Answer 1

Given, In $\Delta ABC,DE||BC$


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

Construction : Draw $EM\perp AB$ and $DN\perp AC$. Join B to E and C to D.

Proof: In $\Delta ADE$ and $\Delta BDE$

$\frac{ar\left(\Delta ADE\right)}{ar\left(\Delta BDE\right)}=\frac{\frac{1}{2}\times AD\times EM}{\frac{1}{2}\times DB\times EM}=\frac{AD}{DB}$ . . . (i)
[Area of $\Delta =\frac{1}{2}\times \text{base}\times \text{corresponding altitude}$]

In $\Delta ADE$ and $\Delta CDE$

$\frac{ar\left(\Delta ADE\right)}{ar\left(\Delta CDE\right)}=\frac{\frac{1}{2}\times AE\times DN}{\frac{1}{2}\times EC\times DN}=\frac{AE}{EC}$ . . . (ii)
Since, $DE||BC$ [Given]
$\therefore ar\left(\Delta BDE\right)=ar\left(\Delta CDE\right)$ . . . (iii)
[$\Delta s$ on the same base and between the same parallel sides are equal in area]

From eq. (i), (ii) and (iii)

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