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

Ans:

Given: ABCD is a parallelogram and line AD is extended to point E. Line BE intersects CD at point F

To Prove: △ ABE ~ △ CFB

Construction: We draw a parallelogram ABCD.

Next we extend line AD to point E

Then we connect B with E

F is the intersection point on CD

Solution:

(This question can be solved by two methods. Both are explained below. Only one solution to be written in exam, not both.)

Method 1:

∵ ABCD is a parallelogram:

∴ ∠ BAE  = ∠ BCF      (Opposite angles of parallelogram)

Next, Since AE ǁ BC and line CE cuts it,

∴ ∠ AEB  = ∠ FBC      (Alternate interior angles)

∴ By AA similarity criterion:

△ ABE ~ △ CFB …. Hence Proved !

Method 2:

Let’s make a diagram to start solving:

In △ ABE, DF ǁ AB, then by BPT theorem,

∴

∴

∴ 

∴ 

∵ ABCD is a parallelogram,

∴ AD = BC

∴ …………. (i)

In the given diagram: AE ǁ BC and line CE cuts it,

∴ ∠ AEB  = ∠ FBC      (Alternate interior angles)

Now by SAS similarity criterion, we get:

△ ABE ~ △ CFB …. Hence Proved !

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