Question

In a parallelogram ABCD, AE=EF=FC. Prove that i. DE is parallel to FB ii. DE=FB iii. DEBF is a parallelogram

Answer 1

ABCD is a parallelogram
To prove that De parallel to FB
AE=EF=FC(given})
so DE || FB(By Vertically opposite side)
proved
DEBF is a parallelogram
because DEBF is in parallelogram property
proved
DE=FB we have to prove
AE=EF(given)
E=F(given)
DE=FB
proved

Answer 2

Answer:

i) ∠AE=∠EF=∠FC ----(given)

Step-by-step explanation:

In ΔADE and ΔDFC,

AD=BC----(opp sides of a parallelogram are equal)

AE=FC ---(given)

DE=FB---(as AE=FC=EF and AC bisects ∠A and  ∠C)

∴ΔDAE ≅ΔBFC

Now DE=FB

∴DE//FB

HENCE PROVED

ii)DE=FB (proved above(i))

iii)DA =AB ----(proved in question (i))

and DC=AB

and OA//BC,AB//DC

∴ABCD is a parallelogram