Question

In a parallelogram ABCD , e is any point on side BC. Diagonal BD and AE intersect at P. Prove that dp*ep=pb*pa

Answer 1

Consider ΔAFD and ΔBEF
∠ABD and ∠CBD are alternate angles

∠ADF = ∠EBF
∠DFA = ∠BFE
as they are vertically opposite

so ΔAFD is congurent to ΔEFB

DF/ FA = BF/FE
or
DF x FE= BF x FA