Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
Answer :
XY || BC = EY || BC ...(Given)
Also, BE || AC = BE || CY
\(\therefore\) we can say that, EBYC is a parallelogram.
Similarly,
XY || BC = XF || BC
Also, FC || AB = FC || XB
\(\therefore \) BCFX is a parallelogram.
Parallelograms EBCY and BCFX are on the same base BC and between the same parallels BC and EF.
\(\therefore \) Area (EBCY) = (\(\frac{1}{2} \) ) Area (BCFX) ...(i)
Now, considering parallelogram EBYC and \(\triangle{AEB}\),
We can say that, these lie on the same base BE and are between the same parallels BE and AC.
\(\therefore \) Area (\(\triangle{AEB}\)) = (\(\frac{1}{2} \) ) Area (EBYC) ...(ii)
Also, parallelogram BCFX and \(\triangle{ACF}\) are on the same base CF and between the same parallels CF and AB.
Thus, Area (\(\triangle{ACF}\)) = (\(\frac{1}{2} \) ) Area (BCFX) ...(iii)
So, from eq.(i), (ii) and (iii), we get,
Area (\(\triangle{AEB}\)) = (\(\triangle{ACF}\))
Hence, proved.