Q12 of 40 Page 8

ABCD is a parallelogram; AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC.

To Prove: BF = BC

Given: AD is produced to E, DE = DC


EC is produced to meet AB in F.


Concept Used:


Opposite sides of a parallelogram are equal.


ASA Congruence: The ASA rule states that. If two angles and the included side of one triangle are equal to two angles and included a side of another triangle, then the triangles are congruent.


Diagram:



Explanation:


In Δ ACE, D, and O are mid points of AE and AC respectively.


∴ DO||EC


⇒ OB||CF


⇒ AB = BF ….(i)


⇒ DC = BF [AB = DC as ABCD is a parallelogram]


In Δ’s EDC and CBF, we have


⇒ DC = BC


⇒ EDC = CBF


and ECD = CFB


So, by ASA congruence criterion, we have


ΔEDC ≅ ΔCBF


⇒ DE = BC


⇒ DC = BC


⇒ AB = BC


⇒ BF = BC [AB = BF from (i)]


Hence, Proved.


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