Question

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Answer 1

Answer:

Taken ∆Abc and ∆Def

<abc =<dfe ( Ab parallel to fe and bf is a transversal so )

<c=<d given

AC =DE given

so ∆s are congruent by aas property

by cpct Ab=Fe

and bc =df

cutcd from both side

then

BD=CF

Answer 2

Answer:

Given:

AC=DE

<EDF = <ACB

To Find:

AB=FE

BD=CF

Solution :

consider triangle ABC, FED

AC =DE

<EDF =<ACB

<DFE = <CBA(angles opposite to equal sides are equal )

By SAA criteria

triangle ABC is congruent to triangle FED

By CPCT

AB =FE

BC = DF

BD+DC = CF+DC

BD =CF

Hence proved.......

Hope this helps you. . . . . . . .

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