1

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?

enter image description here

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). this is the whole proof as requested.

Jamminermit
  • 1,923

2 Answers2

1

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.

Jamminermit
  • 1,923
0

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 $\angle BCD = \frac{\text{arc } BD}{2}$.

Toby Mak
  • 16,827
  • Ok, but how you would go about angle ABD. Additionally, the proof just mentions this is passing and doesn’t go into great detail, so it can’t be too complicated. – Jamminermit Jul 30 '19 at 13:51
  • It's not complicated at all: I only used basic theorems. I answered when you didn't have the full proof, so how was I supposed to know that you only needed to prove $\angle BCD = \angle ABD$? – Toby Mak Jul 31 '19 at 01:34
  • Yeah that’s fine, you answered my original question that I needed. In terms of the simplicity of the proof, it is relatively straight forward but uses knowledge that wouldn’t be expected from where I got this proof (specifically that angle size is radius * arc length. – Jamminermit Jul 31 '19 at 09:55