Question

E is a point on the side AD produced of a parallelogram ABCD and BE Intersects CD at F. Show that ∆ABE ~ ∆CFB.

Answer 1

Data: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F.

To Prove: ∆ABE ~ ∆CFB

In □ABCD Adjacent angles are equal.

Let ∠DAB = ∠BCD = 70°

∠DAB = ∠EAF = 70° (∵ Corresponding angle)

In ∆EDF, ∠DEF = 30° then,

∠EFD = 80°.

∠EFD = ∠BFC = 80° (vertIcally opposite angles)

In ∆FBC, ∠FBC = 30°,

Now in ∆ABE and ∆CFB,

∆EAB = ∆BCF = 70°

∆AEB = ∆FBC = 30°

∆ABE = ∆BFC = 80°

∴ Similarity criterion for ∆ is A.A.A.

∴ ∆ABE ~ ∆CFB