Answer:
The proof is explained step-wise below :
Step-by-step explanation :
For better understanding of the solution see the attached figure :
Given : ABCD is a Parallelogram ⇒ AB ║ DC and AD ║ BC
Now, F lies on the extension of DC. So, AB ║ DF
To Prove : ΔABE is similar to ΔFCE
Proof :
Now, in ΔABE and ΔFCE
∠ABE = ∠FCE ( alternate angles are equal )
∠AEB = ∠FEC ( Vertically opposite angles )
So, by using AA postulate of similarity of triangles
ΔABE is similar to ΔFCE
Hence Proved.
