Please help me as fast as you can. thanks

Step-by-step explanation:

In triangle DEF, we have:

Given:

<EDF=56

<EFD=84

So, <DEF =180 - 56 - 84 =40 (sum of triangle angles is 180)

____________

DE is a midsegment of triangle ACB

( since CD=DA(given)=>D is midpoint of [CD]

and BE = EA => E midpoint of [BA] )

According to midsegment Theorem,

(DE) // (CB) "//"means parallel

and DE = CB/2 = FB =CF

___________

DEBF is a parm /parallelogram.

Proof: (DE) // (FB) ( (DE) // (CB))

AND DE = FB

Then, <EBF = <EDF = 56

___________

DEFC is parm.

Proof: (DE) // (CF) ((DE) // (CB))

And DE = CF

Therefore, <DCF = <DEF =40

___________

In triangle ACB, we have:

<CAB =180 - <ACB - <ABC =180 - 40 - 56 =84(sum of triangle angles is 180)

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step-by-step explanation:

i think because what else can be a sample of every member