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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=

Sagot :

<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)