AB = 5 sin ∠B = 3/5

BG = AG = 3

Because AD = 2

BC = 4 +2 +4 = 10

(2)B =∠C

△ Abe ∠BAE =∠CEF, that is △ECF.

AEB =∠AEF =∠ Conventional Armed Forces in Europe

AE / / CF

Then the quadrilateral AECF is a parallelogram.

CE = AD = CF = AE = 5，

Point f coincides with point d, which does not meet the meaning of the question and is rounded off.

∠BEA =∠CEF, △ Abe is almost the same ∠AEB =△FCE.

AEF =∠CEF = 60

Then EG = square root 3

AB / CF = BE / CE

5/CF =(4+ root number 3)/(square root of 6-3)

CF =( 135-50 root number 3)/ 13

(3)y =( 10x ^ 2- 120 x+400)/(x ^ 2- 16x+39)

Domain 4