In triangle ABC, angle B = 120 degrees and angle A = 30 degrees.

In triangle ABC, angle B = 120 degrees and angle A = 30 degrees. Point D belongs to side AC, and angle BDC is obtuse. Prove that AB is greater than BD

In triangle ABC, angle ACB = (180 – 120 – 30) = 30.

Since point D is located on the segment AC, then in triangle ABD the angle ADB will always be greater than 30 if point D does not coincide with point C.

Then the angle ADB> BAD, and hence AB> BD, which was required to be proved.