i) $$\angle ABC = 120^\circ$$. ii) $$\triangle ACD$$ is equilateral, iii) B and D are on opposite sides of $$\overleftrightarrow{AC}$$. Prove that a) $$\overleftrightarrow{BD}$$ bisects $$\angle ABC$$ and b) $$\overline{BD}$$ is in length equal to the sum of lengths of $$\overline{AB}$$ and $$\overline{BC}$$.

PLACEHOLDER ANSWER - this is a proof question. Since $$\angle ABC + \angle ADC = 120^\circ + 60^\circ = 180^\circ$$ and B, D are on opposite sides of AC, ABCD is cyclic. a) Angles in the same segment give $$\angle ABD = \angle ACD = 60^\circ$$ and $$\angle DBC = \angle DAC = 60^\circ$$, so BD bisects $$\angle ABC$$. b) By Ptolemy's theorem, $$AC \cdot BD = AB \cdot CD + BC \cdot AD$$, and since $$AC = CD = AD$$, $$BD = AB + BC$$.