Question

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

• A. 4
• B. $3\sqrt{3}$
• C. $2\sqrt{3}$
• D. 2

Answer 1

The correct option is A 4

Let $\angle BAD=\angle DAC=x$ (AD is angle bisector)

Now, $\angle BCD=\angle BAD=x$ (Angles in same segment)

$\Rightarrow \angle BAC=\angle ACB=2x$ (Equilateral triangle)

Now $\angle DOC=2x$

In $△$ DOC, OD = DC

In $△$ BDC, BD = DC

And $BD+DC=4\Rightarrow BD=DC=2$

$\Rightarrow OD=2(\because OD=DC)$

$\therefore$ Diameter $=2\left(OD\right)=4$