Question

In ∆ABC, DE || BC; DC and EB intersect at F.

Prove that
ar(∆ADE)/ar(∆ABC)=
ar(∆FDE)/ar(∆FBC)
​

Answer 1

in training ADE and ABC

DE // BC

so ADE ~ ABC

${ (\frac{de}{bc} )}^{2} = \frac{ar(ade)}{ar(abc)}$

also in triangle DEF and BFC

angle DFE = angle BFC

angle DEF = angle CBF

so using aa similaraly

DEF ~ CBF

${ (\frac{de}{bc} )}^{2} = \frac{ar(def)}{ar(cbf)}$

from both the above Formula

They are equal

$\frac{ar(ade)}{ar(abc)} = \frac{ar(def)}{ar(cbf)}$

hence proved