Answer:
<DEF = 40
<EBF = <EDF = 56
<DCF = <DEF =40
<CAB = 84
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)
