Question

In the figure, ABCD and AEFD are two parallelograms. Prove that
$\left(i\right)PE=FQ\left(ii\right)ar\left(△APE\right):ar\left(△PFA\right)=ar\left(△QFD\right):ar\left(△PFD\right)\left(iii\right)ar\left(△PEA\right)=ar\left(△QFD\right).$

Answer 1

Given: Two ||gm ABCD and ||gm AEFD are on the same base AD. EF is produced to meet CD at Q. JD AF and PD and PD also
To Prove:
$\left(i\right)PE=FQ\left(ii\right)ar\left(△APE\right):ar\left(△PFA\right)=ar\left(△QFD\right):ar\left(△PFD\right)\left(iii\right)ar\left(△PEA\right)=ar\left(△QFD\right).$

Proof:
(i) In $△AEPandDFQ,$
AE = DF ( Opposite sides of a ||gm)
$\angle AEP=\angle DFQ$ (Corresponding angles)
$\angle APE=\angle DQF$ (Corresponding angles)
$\therefore △AEP\cong △DFQ\therefore PE=QF$
(ii) and $ar\left(△AEP\right)=ar\left(△DFQ\right)$
(iii) $\because △PFAand△PFD$ are on teh same base PF and between the same parallels
$\therefore ar\left(△PFA\right)=ar\left(△PFD\right)$
From (i) and (ii) ,
$\frac{ar\left(△AEP\right)}{ar\left(△PFA\right)}=\frac{ar\left(△DFQ\right)}{ar\left(△PFD\right)}$