Question

А
23. In the given figure, XY || BC, BE || CA and FC || AB.
Prove that : ar (AABE) = ar (AACF).
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Answer 1

Answer ⤵️⤵️

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

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

Given BE∥CA and CF∥BA

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

So  EY∥BC and EB∥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(ΔABE)=21areaofparallelogram(BCAE)..................(2)

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

 Then area(ΔACF)=21areaofparallelogram(BCAF)..................(3)

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

areaofΔABE=areaofΔACF [henceproved]