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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 :

[tex]\\ \rm\hookrightarrow AE\cong FC[/tex]

So parallelogram BGAE and HDFC are equal.

  • GE=FC

And

  • BG=HD

As BG=HD

  • GC=AH

Hence

[tex]\\ \rm\hookrightarrow \triangle GEC\cong \triangle HFA(SSA)[/tex]

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