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I am stuck on the following question:

Prove that each diameter is twice as long as each radius.

I drew a circle, with center O and diameter AB. Is there a theorem that could help me say that congruent segment AO and BO add up to form segment AB?

Or is there some other way to prove this?

I would really like it if anyone could give me a hint about this.

Thank you.

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Hint:

A circle is (in part) defined by having an equal distance from its center to its edge for all points on its edge, i.e. it has a constant radius. So then what's the distance from the edge to the center to the edge again? (And does that sound related to your definition of a diameter at all?)

You could also prove this pretty easily by contradiction: "Suppose $d \neq 2r$. Then..." I'll leave that to you.

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  • $\begingroup$ Okay I think I can prove the first one. However, I am not sure how to continue the 2nd option (proof by contradiction), could you continue it a little? $\endgroup$ – Michael Ferashire Silva Oct 29 '13 at 7:38
  • $\begingroup$ So if $d \neq 2r$, then either $d > 2r$ or $d < 2r$. If $d > 2r$, and we have the diameter as a chord between points $A,B$ on the circle, and call $C$ the center, then we can draw a radius line $A \rightarrow C$ and a radius line $B \rightarrow C$, but by our assumption that won't be equal to the diameter? The other case uses a similar sort of logic - you just have to show that if you suppose it is true, then it leads to an absurdity. $\endgroup$ – Newb Oct 29 '13 at 7:42
  • $\begingroup$ Okay thank you very much Newb :) $\endgroup$ – Michael Ferashire Silva Oct 29 '13 at 8:00
  • $\begingroup$ You're welcome! :-) $\endgroup$ – Newb Oct 29 '13 at 15:34

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