Question

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

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

Answer 1

Answer: (B)

Given: l||m||n and we have two transversal p and q, there cut A,B,C and D,E,F.

By Construction, we take point O at the intersection of both transversed line p and q.

and $FX\perp OCandCY\perp OF$.Join $AF,BF,DC$ and $EC.$

$Since,FX\perp OC,SoFX$is height of$△OCF\&△BCF.$

$ar\left(△OCF\right)=\frac{1}{2}\times OC\times FX$

$ar\left(△BCF\right)=\frac{1}{2}\times BC\times FX$

$\therefore \frac{ar\left(△BCF\right)}{ar\left(△OCF\right)}=\frac{BC}{OC}-\left(1\right)$

Similarily,Since $CY\perp OF$,So $CY$ is height of$△OCF\&△ECF,$

$\therefore \frac{ar\left(ECF\right)}{ar\left(OCF\right)}=\frac{EF}{OF}-\left(2\right)$

But$△BCFand△ECF$ are on the same base $CF$ b/w same parallel straight lines BC& CF {m||n}

$△BCF=△ECF-\left(3\right)$

$\therefore \frac{BC}{OC}=\frac{EF}{OF}-\left(A\right)$

$Since,FX\perp OC,FX$ is height of$△OCF\&△ACF$

$ar\left(△OCF\right)=\frac{1}{2}\times OC\times FX$

$ar\left(△ACF\right)=\frac{1}{2}\times AC\times FX$

$\frac{ar\left(ACF\right)}{ar\left(OCF\right)}=\frac{AC}{OC}=\frac{AB+BC}{OC}-\left(4\right)$

$Since,CY\perp OF,SoCYisheightof△OCF\&△DCF.$

$\frac{ar\left(DCF\right)}{ar\left(OCF\right)}=\frac{DE+EF}{OF}$

$But△ACF\&△DCF$ are on same base CF and b/w same parallel straight line,

$Since,ADandCF\left\{AD||CFasl||n\right\}$

$△ACF=△DCF-\left(6\right)$

$\frac{AB}{OC}+\frac{BC}{OC}=\frac{DE}{OF}+\frac{EF}{OF}$

$⟹\frac{AB}{OC}=\frac{DE}{OF}-\left(B\right)\{\because \frac{EF}{OF}=\frac{BC}{OC}\}$

$\therefore \frac{\frac{AB}{OC}}{\frac{BC}{OC}}=\frac{\frac{DE}{OF}}{\frac{EF}{OF}}$

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