Question

If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it ....... the two sides in the same ratio.

• A. Subtracts
• B. Adds
• C. Multiplies
• D. Divides

Answer 1

The correct option is D Divides

Correct option is $\text{D}$.

If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.

Given that: In a $\Delta ABC$, line $DE$ is parallel to $BC$ and line $DE$ intersects $AB$ at $D$ and $AC$ at $E$.

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

Construction: Join $BE$ and $CD$ and draw a perpendicular $EM$ at $AB$ and $DN$ at $AC$.

Proof

$\frac{ar\left(\Delta ADE\right)}{ar\left(\Delta BDE\right)}=\frac{\frac{1}{2}\times EM\times AD}{\frac{1}{2}\times BD\times EM}$

$\Rightarrow \frac{ar\left(\Delta ADE\right)}{ar\left(\Delta BDE\right)}=\frac{AD}{BD}...\left(1\right)$

Similarly,

$\frac{ar\left(\Delta ADE\right)}{ar\left(\Delta CDE\right)}=\frac{\frac{1}{2}\times AE\times DN}{\frac{1}{2}\times EC\times DN}$

$\Rightarrow \frac{ar\left(\Delta ADE\right)}{ar\left(\Delta CDE\right)}=\frac{AE}{BC}...\left(2\right)$

But $ar\left(\Delta BDE\right)=\mathrm{ar}\left(\Delta CDE\right)$ (triangle on same base $DE$ and between the same parallels $DE$ and $BC$ )

Thus, equation (2) becomes,

$\frac{ar\left(\Delta ADE\right)}{ar\left(\Delta BDE\right)}=\frac{AE}{EC}...\left(3\right)$

From equation $\left(2\right)$ and $\left(3\right)$, we have,

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

Hence Proved.