matteop7 matteop7

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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Аноним Аноним

06-03-2022

[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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