Question

Area $\left(\Delta AED\right)=2\times area\left(\Delta CEX\right)$.

Answer 1

Consider the figure,

$BE\parallel AD$ [ Opposite side of $\parallel$gm ]

$\Rightarrow AB=BX$ (given)

$\Rightarrow DE=EX$

$\Rightarrow BE=\frac{AD}{2}$ [ By mid point theorem : segment joining mid points of $2$ sides of a triangle is parallel to the $3rd$ side and equal half of it ]

$\Rightarrow AD=BC$ $\to BE=\frac{BC}{2}$

$EX$ is a median of $△CBX$

$\Rightarrow ar\left(△EBX\right)=ar\left(△CEX\right)⟶\left(1\right)$

$ar\left(△EBA\right)=ar\left(△EBX\right)⟶\left(2\right)$

$ar\left(△DAX\right)=\frac{1}{2}\times AX\times h$

$ar\left(△EAX\right)=\frac{1}{2}\times AX\times \frac{h}{2}$

$\Rightarrow ar\left(△DAX\right)=2ar\left(△EAX\right)$

$ar\left(△DAX\right)=2\times 2x=4x$

$ar\left(△AED\right)=ar\left(△DAX\right)-\{ar\left(△AEB\right)+ar\left(△EBX\right)\}$

$ar\left(△AED\right)=4x-2x=2x$

$ar\left(△AED\right)=x$

$\therefore ar\left(△AED\right)=2ar\left(△CEX\right).$

Hence, answer solved.