Question

XY is a line parallel to side BC of a triangle ABC. If BE||AC and CF||AB meet XY at E and F respectively, show that area of $△ABE=$area of $△ACF$

Answer 1

$\because XY\parallel BC$(given)

and $CF\parallel BX$(given) $\because CF\parallel AB$

$\therefore BCFX$ is a parallelogram.

A quadrilateral is a parallelogram if its opposite sides are parallel.

$\therefore BC=XF$

Opposite sides of a parallelogram are equal$\Rightarrow BC=XY+YF$ ....$\left(1\right)$

Again, $\because XY\parallel BC$(given)

$\therefore BCYE$ is a parallelogram.

A quadrilateral is a parallelogram if its opposite sides are parallel

$\therefore BC=YE$

$\Rightarrow BC=XY+YE$ ...$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

$XY+YF=XY+XE$

$\Rightarrow YF=XE$ ...$\left(3\right)$

$\because △AEX$ and $△AYF$ have equal bases ($\because XE=YF$) on the same line $EF$ and have a common vertex $A$

$\therefore$ Their altitudes are also the same.

$\therefore$area$\left(△AEX\right)=$area$\left(△AFY\right)$ ....$\left(4\right)$

$\because △BEX$ and $△CFY$ have equal bases ($\because XE=YF$) on the same line $EF$ and between the same parallels $EF$ and $BC$$

$\therefore$ area$\left(△BEX\right)=$area$\left(△CFY\right)$ ....$\left(5\right)$

Two triangles on the same base(or equal bases) and between the same parallels are equal in area.

Adding the corresponding sides of $\left(4\right)$ and $\left(5\right)$, we get

area$\left(△AEX\right)+$area$\left(△BEX\right)=$area$\left(△AFY\right)+$area$\left(△CFY\right)$

$\Rightarrow$ area$\left(△ABE\right)=$area$\left(△ACF\right)$

Hence proved.