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Ghella [55]
2 years ago
13

See Problem

Mathematics
1 answer:
Kazeer [188]2 years ago
3 0

Answer:

its question 4

Step-by-step explanation:

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Please help me as fast as you can. thanks
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Answer:

<h2><DEF = 40</h2><h2><EBF = <EDF = 56</h2><h2><DCF = <DEF =40</h2><h2><CAB = 84</h2>

Step-by-step explanation:

In triangle DEF, we have:

<u>Given</u>:

<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.

<u>Proof</u>: (DE) // (FB) ( (DE) // (CB))

AND DE = FB

Then, <EBF = <EDF = 56

___________

DEFC is parm.

<u>Proof</u>: (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)

HOPE \:  THIS \:  HELPS.. GOOD  \: LUCK!

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