Question

Let ABC be an equilateral triangle. The bisector of ∠BAC meets the circumcircle of ABC in D. Suppose DB + DC = 4. The diameter of the circumcircle of triangle ABC is

Answer 1

Let ∠BAD = ∠DAC = x (AD is angle bisector)
Now, ∠BCD = ∠BAD = x (Angles in same segment)
⇒ ∠BAC = ∠ACB = 2x (Equilateral triangle)
Now ∠DOC = 2x
In △DOC, OD = DC
In △BDC, BD = DC
And BD + DC = 4 ⇒ BD = DC = 2
⇒ OD = 2 (∵ OD = DC)
∴ Diameter = 2(OD) = 4