Parallelogram ABCD is given. Draw line EF so that it goes through the vertex A. Point E lies on the side BC and point F lies on
the extension of DC. Prove that ABE is similar FCE
1 answer:
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.
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And there's our answer. Hopefully that's helpful! :)