Question

State true or false:

In the parallelogram $ABCD$ , the side $AB$ is produced to the point $X$, so that $BX=AB$ . The line $DX$ cuts $BC$ at $E$, then

$Area\left(△AED\right)=2\times Area\left(△CEX\right)$

• A. True
• B. False

Answer 1

The correct option is A True

Since, $BE\parallel AD$ ....... (Opposite sides of parallelogram)

$AB=BX$ ..... (Given)

$\Rightarrow DE=EX$

By mid point theorem: Segment joining mid points of 2 sides of a triangle is parallel to the 3rd side & half of it

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

But $AD=BC$

$BE=\frac{BC}{2}$

i.e, $EX$ is a median of triangle $CBX$

So, $area\left(△EBX\right)=area\left(△CEX\right)$ ..... $\left(1\right)$ (As a median divides the triangle into $2$ triangles of equal area)

Similarly,

$area\left(△EBA\right)=area\left(△EBX\right)$ .... $\left(2\right)$ (Since median $EB$ divides $△EAX$ into $2$ triangles $EBA$ and $EBX$ of equal area)

By (1) & (2) let all these 3 triangles’ area be X unit²

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

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

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

$\Rightarrow$ $area\left(△DAX\right)=2X\times 2X=4X$

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

$area\left(△AED\right)=4X-2X=2X$

and $area\left(△CEX\right)=X$

Hence, $Area\left(△AED\right)$ $=2\times Area\left(△CEX\right)$