Solution:

① Point D is located on the circle O for the following reasons:

Remember that BC and circle o intersect at d',

In triangle ABD', BD' = AB * COS 30 = 2 √ 3.

D is the midpoint of BC line.

∴BD= 1/2BC=2√3

∴BD=BD'

That is, d and d' coincide.

Point d is on the circle o.

Prove:

② Connecting outer diameter

D is the midpoint of BC line.

O is that midpoint of the diameter AB.

∴OD is the center line of △ABC.

∴OD‖AC

You are ∵DE⊥AC.

∴∠CED=90

∴∠EDO=∠CED=90

Starting from ①, point D is on circle O.

∴OD is the radius of the circle o.

The straight line DE is tangent to the circle O.