Question

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

• A. perpendicular
• B. parallel
• C. ${60}^o$
• D. ${30}^o$

Answer 1

The correct option is B parallel

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

This is called as basic proportionality theorem.

$\text{Explanation:}$

Construction:

$ABC$ ia a triangle. $DE\parallel BC$ and $DE$ intersects $AB$ at $D$ and $AC$ at $E$.

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

To prove:

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

Proof:

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

$ar\left(ADE\right)=\frac{1}{2}\times AE\times DN=\frac{1}{2}\times AD\times EM$

$ar\left(DEC\right)=\frac{1}{2}\times EC\times DN$

Hence,

$\frac{ar\left(ADE\right)}{ar\left(DEC\right)}=\frac{AE}{EC}$ ...(1)

And, $\frac{ar\left(ADE\right)}{ar\left(BDE\right)}=\frac{AD}{BD}$ ...(2)

Triangles $BDE$ and $DEC$ are on the same base, i.e. $DE$ and between same parallels, i.e. $DE$ and $BC$.

Hence, $ar\left(BDE\right)=ar\left(DEC\right)$

Dividing eq(1) by eq(2), we get

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

Hence proved.