Question

XY is a line parallel to side BC of a triangle ABC, If $BE\parallel AC$ and $CF\parallel AB$ meet XY at E and F respectively,show that ar(ABE)=ar(ACF)

Answer 1

$\Delta ABC$

where, $XY||BC$ & $BE||AC$ & $CF||AB$

Let XY intersect AB and BC at M and N respectively

Given : $XY||BC$

$\Rightarrow EN||BC$

Also given $BE||AC$

$\therefore BE||CN$

In $BCNE$

both opposite sides are parallel

$\therefore BCNE$ is parallelogram

Parallelogram $BCNE$ and $\Delta AEB$ lie on same base BE and are between same parallel $BE\&AC$

$\therefore Ar\left(\Delta ABE\right)=\frac{1}{2}Ar\left(BCNE\right)$ __(1)

Similarly , $BCFM$ is parallelogram

$\therefore Ar\left(\Delta ACF\right)=\frac{1}{2}Ar\left(BCDM\right)$ __(2)

Parallelogram $BCNE$ and $BCFM$ are on

same base BC and between same parallel BC and EF

$\therefore Ar\left(BCNE\right)=Ar\left(BCFM\right)$ __(3)

from eq 1, 2, 3

$Ar\left(ABE\right)=Ar\left(ACF\right)$