Let triangle ABC and a point D on AC such that BD = DC = 3. If AD = 6 and ∠ ACB = 30◦, calculate ∠ ABD.
Triangle BDC is a 30-30-120 isosceles triangle, then BC=3*sqrt(3).
Let point E be the base of the altitude from B onto side AC.
BEC is a 30-60-90 right triangle. Then BE=(3/2)*sqrt(3)
BED is a 30-60-90 right triangle. Then EC=9/2.
Then DE=3/2 and AE=9/2
AEB is a right triangle with legs AE and EB. AE/EB=sqrt(3) so it is a 30-60-90 triangle.
Angle ABE=60 and angle EBD=30, so angle ABD=90 degrees.