In the given figure, if PB || CF and DP || EF, then ADDE=

Given: PB||CF and DP||EF. AB = 2 cm and AC = 8 cm.

To find: AD: DE

According to BASIC PROPORTIONALITY THEOREM, if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In ∆ACF, PB || CF.

$\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{AP}}{\mathrm{PF}}$

$\frac{\mathrm{AP}}{\mathrm{PF}}=\frac{2}{8-2}$

$\frac{\mathrm{AP}}{\mathrm{PF}}=\frac{2}{6}$

$\frac{\mathrm{AP}}{\mathrm{PF}}=\frac{1}{3}$.....(1)

Again, DP||EF.

$\frac{\mathrm{AD}}{\mathrm{DE}}=\frac{\mathrm{AP}}{\mathrm{PF}}$

$\frac{\mathrm{AD}}{\mathrm{DE}}=\frac{1}{3}$

Hence we got the result option $(b)$.