Question

Question 8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB? Can it done by AAA similarity?If no then give reason.

Answer 1

Answer:

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Given: A parallelogram ABCD where is the point of the side AD produced & BE intersects at F.

To prove: ΔABE ∼ ΔCFB

Proof :

In parallelogram ABCD,

Opposite angles are equal.

Hence, angle A = angle C......(1)

Also in parallelogram ABCD opposite sides are parallel.

AD||BC

Now, since AE is AD extended

AE||BC

& BE is the traversal.

Hence, Angle AEB = Angle CBF(alternate angles).......(.2)

Now, in triangle ABC & Triangle CFB

angle A = angle C (from 1)

angle ABE = angle CBF (from 2)

Hence, ΔABE ∼ ΔCFB ( AA similarity criterion)

(proved)

At last, it can be done by similarity criterion.

Hope it’s helpful....... ☺