Solution

Let O be the circumcenter of ΔABC, K the midpoint of BC. Then in right triangle OCK, COK = BAC, which I'll also denote A. Thus

The altitude of the corresponding circular segment is

Similarly, in obvious notations,

Therefore,

Now taking into account that

easily yields the desired result.

Remark

If we denote α = COK = A, and assume the circumradius of ΔABC equals 1, then CK = sin(α) and OK = cos(α). If S is the midpoint of arc OK below K, then KS also is a named quantity, viz., versine (versed sine) of α:

As we saw,