Question

In the figure, XY is a line parallel to BC is drawn through A. If BE || CA and CF || BA are drawn to meet XY at E and F respectively. Show that $area\left(△ABE\right)=area\left(△ACF\right)$.

Answer 1

Given in figure, XY is a line parallel to BC is drawn through A.

If $BE\parallel CA$ and $CF\parallel BA$ are draw to meet XY at E and F respectively.

Given $BE\parallel CA$ and $CF\parallel BA$

Then $EF\parallel BC$ because of XY parallel to BC and point E, A F, on the line BC

So $EY\parallel BC$ and $EB\parallel CY$ The part of line XY

Then BCAE and BCAF are the parallelograms

The BCAE and BCAF is the parallelogram at on same base BC and

parallel line XY and BC

Then the area of parallelogram (BCAE)=area of a parallelogram(BCAF) ...............(1)

The triangle ABE and parallelogram (BCAE) are on the same base BC and two parallel line BC and EF

Then $area\left(\Delta ABE\right)=\frac{1}{2}areaofparallelogram\left(BCAE\right)$..................(2)

The triangle ACF and parallelogram (BCAF) are on the same base BC and two parallel line BC and EF

Then $area\left(\Delta ACF\right)=\frac{1}{2}areaofparallelogram\left(BCAF\right)$..................(3)

From (1), (2) and (3) we get

$areaof\Delta ABE=areaof\Delta ACF$ $\left[henceproved\right]$