Here's another approach to (2)
Let the four equal angles in the question = a
Since < DEC = < ACB , then EB = BC
And EB = BD + DE = 8 so....
8 = 2 + DE → DE = 6
And < ECD = 180 - 2a
So....using the Law of Sines, we have :
sin DEC / DC = sin ECD / 6
sin (a) / DC = sin(180- 2a)/ 6
sin(a) / DC = sin(2a)/ 6
sin(a)/DC = [2sin(a)cos(a)] / 6
DC /sin(a) = 6 / [2sin(a)cos(a)]
DC = 6sin(a) / [ 2sin(a)cos(a)]
DC = 3sec(a)= EC
And using the Law of Cosines, we have :
(DC)^2 = 2^2 + 8^2 - (2)(16)cos(DBC)
(3sec(a))^2 = 2^2 + 8^2 - (2)(16)cos(DBC)
cos(DBC) = [9sec^2(a) - 68] / [-32] (1)
And applying it again, we have :
(EC)^2 = 2^2 + 8^2 - (2)(16)cos(DBC)
(3sec(a))^2 = 8^2 + 8^2 - 2(64)cos(DBC)
cos(DBC) = [ 9sec(a)^2 - 128] / [-128] (2)
Equate (1), (2)
[9sec^2(a) - 68] / [-32] = [(9sec(a)^2 - 128] / [-128]
4[ 9sec^2(a) - 68] = 9sec^2(a) - 128
36sec^2(a) - 272 = 9sec^2(a) - 128
27sec^2(a) = 144
sec^2(a) = 144/27 = 16/3
cos^2(a) = 3/16
cos(a) = sqrt(3/16)
arccos(sqrt(3)/4) = a = 64.34° = m< ACB
So m< BAC = [180 - 2a] = [180 - 2(64.34)] ≈ 51.32°
And using the Law of Sines once more, we have that :
AB/sin(ACB) = BC/ sin(BAC)
AB /sin(64.34) = 8/ sin(51.32}
AB = 8*sin(64.34)/sin(51.32) = about 9.237
