In a parallelogram ABCD, from vertex D, a line DE is drawn bisects BC at H and meeting AB (extended) at E. From vertex C, a line CF is drawn bisecting AD and G and meeting AB (extended) at F. Lines DE and CF intersect at P. If the area of parallelogram ABCD is 32 sq. cm, then what is the area (in sq. cm) of triangle PEF?
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In △
′
sEAD and DCF, we have
∠1=∠2 [∵AB∣∣DC∴ Corresponding angles are equal]
∠3=∠4 [∵AD∣∣BC∴ Corresponding angles are equal]
Therefore, by AA-criterion of similarity, we have
△EAD∼△DCF
⇒
DC
EA
=
CF
AD
=
FD
DE
⇒
DC
EA
=
CF
AD
⇒
AE
AD
=
CD
′
sEAD and EBF, we have
∠1=∠1 [Common angle]
∠3=∠4
So, by AA-criterion of similarity, we have
△EAD∼△EBF
⇒
EB
EA
=
BF
AD
=
EF
ED
⇒
EB
EA
=
BF
AD
⇒
AE
AD
=
BE
FB
From (i) and (ii), we have
AE
AD
=
BE
FB
=
CD
CF
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