Well, here's a start!
A = angle A : so angle C = 100 - A which leaves angle BDC = 40 + A
a = BC, b = AC, c = AB, d = BD : so AD = a + d
Insert point E on AD such that DE = d, which leaves AE = a :
so we now have isosceles triangle BDE, equal sides = d, equal angles = (40 + A) / 2.
Insert point F on AB, such that EF is parallel to BC :
so we have triangle AFE, similar to triangle ABC
I think that should lead you to the solution: give it a shot...