Stuck on Circles question on tangents

Question

Here the length of AO is equal to diameter of circle. AB and AC are tangents from A.

The triangle ABC has to be proved equilateral. I put it in geogebra and it was indeed equilateral. I can't find where to start the proof. Any help is greatly appreciated.

Answer 1

$$\triangle ABO \sim \triangle BMO,$$ since $\angle ABO=\angle BMO = 90^\circ$, and $\angle BOA = \angle MOB$.

Then

$$\dfrac{OB}{BM}=\dfrac{OA}{BA};$$ $$\dfrac{R}{BM}=\dfrac{2R}{BA};$$ $$2BM=BA;$$ $$BC=BA.$$ (where $R$ is the radius of the circle).

Answer 2

Angle at tangent point $B$ is $90^0$

$$\sin BOA = \dfrac{BO}{AO} =\dfrac{1}{2}$$

$ BAO = 30^0 , BAC = 60^0,CBA = 90^0 - OBC = 60^0 $, so also for BCA, making triangle ABC equilateral.