**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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