Proving similarity of triangles in the power of point theorem.

Question

Maybe I’m missing something obvious, but why can we say:

Angle BCD = (arc BD)/2?

Similarly why can we say:

Angle ABD = (arc BD)/2?

This is a step in part of a proof for the power of a point theorem (specifically trying to prove that triangle ADB is similar to triangle ABC).

Answer 1

Partial answer:

Let $O$ be the origin. Then we have $\angle BOD = 2 \angle BCD$, or $\angle BCD = \frac{1}{2} \angle BOD$ since the angle at the centre is twice that at the circumference (link).

If we express $\angle BOD$ in radians, we can use the formula $\text{angle size} = \text{radius} \cdot \text{arc length}$. Assuming the radius is $1$, we have $\angle BOD = \text{arc } BD$, so $\angle BCD = \frac{\text{arc } BD}{2}$.

Answer 2

Actually, a simpler answer for this proof simply seems to be to state that angle BCD is equal to angle ABD using the alternate segment theorem, instead of proving both angles are equal to (arc BD)/2.