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06-03-2022
Mathematics

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Given: ABCDis a parallelogram ∠GEC ≅ ∠HFA and AE ≅FC.
Prove △GEC ≅ △HFA.

Given ABCDis a parallelogram GEC HFA and AE FC Prove GEC HFA class=

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10-03-2022

<GEC=<HFA

As

AE=FC
FE=FE(Common in both

So

FE+AE=FE+CE

Cancel FE

A F=E C

Also

<CAH=<GCE(Alternative interior angles)

Henceforth

△GEC ≅ △HFA(ASA)

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