Question

$l$, $m$, $n$ are parallel lines. If line p intersects them at A, B, C and line $q$ at D, E, F then:

• A. $AB=DE$ and $BC=EF$ always
• B. at least one of the pairs AB, DE and BC, EF are necessarily equal.
• C. at least one of the pairs AB, BC and DE, EF are necessarily equal
• D. $\frac{AB}{BC}=\frac{DE}{EF}$

Answer 1

The correct option is D $\frac{AB}{BC}=\frac{DE}{EF}$

Refer figure

Construction:

Join $A\text{to}F\text{such that}AF\text{cuts line}m\text{at}G$

Solution:

In $△ABG\text{and}△ACF$

$\angle B=\angle C$ (alternate angles)

$\angle AGB=\angle AFC$ (alternate angles)

$\angle A$ is common

Thus, $△ABG\text{and}△ACF$ are similar (AAA theorem)

Hence,

$\frac{AC}{AB}=\frac{AF}{AG}$

Subtracting $1$ from each side, we get

$\frac{AC}{AB}-1=\frac{AF}{AG}-1$

$\frac{AC-AB}{AB}=\frac{AF-AG}{AG}$

$\frac{CB}{BA}=\frac{FG}{GA}$...............eq. 1

Similarly in $△FEGand△FDA$

$\frac{DE}{EF}=\frac{AG}{GF}$................eq. 2

Multiplying eq 1 and eq 2, we get

$\frac{CB}{BA}\times \frac{DE}{EF}=\frac{FG}{GA}\times \frac{AG}{GF}$

OR

$\frac{CB}{BA}\times \frac{DE}{EF}=1$

$\frac{CB}{BA}=\frac{EF}{DE}$

OR

$\frac{AB}{BC}=\frac{DE}{EF}$

Ans: Option $D:$ $\frac{AB}{BC}=\frac{DE}{EF}$