Question

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

Answer 1

Answer:

ar (ABE) = ar (ACF)

Step-by-step explanation:

Let's draw the figure as said in the question,

$\because$ XY II BC

$\therefore$ AE II BC

and AB II CF

Hence, AFCB is a parallelogram with AC as a diagonal.

Since, the diagonal of parallelogram divide the parallelogram in two triangles of equal area.

$\therefore$ ar (ABC)= ar (ACF) .......(1)

$\because$ AE II BC

and AE  II BC

Hence, ACBE is also a parallelogram with AB as diagonal

so, ar (ABC)= ar (ABE) .......(2)

from equations (1) and (2), we will get

ar (ABE) = ar (ACF)

proved.