Question

If $l,m$ $n$ are parallel line.
If $P$ interest them at $A,B,$ & $C$ & $Q$ at $DEF$ them
$\frac{AB}{BC}=\frac{DE}{EF}$

• A. True
• B. False

Answer 1

The correct option is A True

$Given:$
$l||m||n\text{and have two transversal p and q cut at A,B,C and D,E,F}$
$\text{By construction m We take point O at the intersection of both transversal p and q}$
$\text{and FX}\perp \text{OC and CY}\perp AF\text{, Join AF ,BF,DC and EC}$
$\text{Since FX}\perp OC\text{so FX is height of}\Delta \text{OCFand}\Delta BCF$
$Ar\left(\Delta OCF\right)=\frac{1}{2}\times OC\times FX$
$Ar\left(\Delta BCF\right)=\frac{1}{2}\times BC\times FX$
$\frac{Ar\left(\Delta OCF\right)}{Ar\left(\Delta BCF\right)}=\frac{BC}{OC}.....\left(i\right)$
$Similarly\mathrm{Sin}ceCY\perp OFsoCYis\text{the height}\text{of}\Delta OCFand\Delta ECF$
$\frac{Ar\left(\Delta ECF\right)}{Ar\left(\Delta OCF\right)}=\frac{EF}{OF}$
$but\Delta BCFand\Delta ECFare\text{on the same base CF}\text{b/w same parallel straight lines BC and CF}$
$\Delta BCF=\Delta ECF.....\left(iii\right)$
$\frac{BC}{OC}=\frac{EF}{OF}......\left(A\right)$
$\mathrm{Sin}ceFX\perp OC,\text{FX is height of}\Delta OCFand\Delta ACF$
$ar\left(\Delta OCF\right)=\frac{1}{2}\times OC\times FX$
$ar\left(\Delta ACF\right)=\frac{1}{2}\times AC\times FX$
$\frac{ar\left(\Delta OCF\right)}{ar\left(\Delta ACF\right)}=\frac{AC}{OC}=\frac{AB+BC}{OC}...\left(iv\right)$
$\mathrm{Sin}ceCY\perp OF\text{so, CY is height of}\Delta OCFand\Delta DCF$
$\frac{ar\left(DCF\right)}{ar\left(OCF\right)}=\frac{DE+EF}{OF}$
$\text{but}\Delta ACFand\Delta DCFare\text{on same base CF and b/w same parallel lines}$
$\text{Since AD and CF}$
$\Delta ACF=\Delta DCF$
$\frac{AB}{OC}+\frac{BC}{OC}=\frac{DE}{OF}+\frac{EF}{OF}$
$\Rightarrow \frac{AB}{OC}=\frac{DE}{OF}.......\left(B\right)$
$\frac{\frac{AB}{OC}}{\frac{BC}{OC}}=\frac{\frac{DE}{OF}}{\frac{EF}{OF}}=\frac{AB}{BC}=\frac{DE}{EF}$
$Hencethegivenstatementistrue$