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kiruha [24]
3 years ago
7

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

Mathematics
1 answer:
tresset_1 [31]3 years ago
7 0

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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