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If $R$ and $r$ be the radii of the circumcircle and incircle of a triangle, then how do I prove by synthetic geometry(i.e. without trigonometry) that $R\ge 2r$?

I am aware of a trigonometric proof but I am not quite sure if I can come up with a synthetic one. In case anyone is interested in the trigonometric proof, I can add it if you ask me to.

Richard Nash
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  • You mean the incircle? – joriki Oct 08 '12 at 11:58
  • @joriki,Oh yes, I do mean incircle. – Richard Nash Oct 08 '12 at 14:17
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    Several of the arguments given in these two questions do not involve trigonometry: http://math.stackexchange.com/questions/170853/why-is-the-inradius-of-any-triangle-at-most-half-its-circumradius/171656#171656 http://math.stackexchange.com/questions/170813/prove-abca-bcab-c-leq-abc-where-a-b-and-c-are-positive-real?lq=1 – Micah Oct 08 '12 at 14:26

1 Answers1

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You can prove something stronger:

$$OI^2=R^2-2Rr$$

Let $ABC$ be your triangle, extend $AI$ until it meets the Circle at $A'$.

Extend $OI$ until it meets the circle at $D$ and $E$. By the power of point to the circle,

$$AI \cdot A'I= ID \cdot IE = (R+OI)(R-OI)= R^2-OI^2 \,.$$

Now, oberve that $ICA'$ is isosceles, by proving that the angles are equal, and obtain $A'I=A'C$.

Last but not least, drop the perpendicular $IC'$ from $I$ onto $AB$, and the diametre $A'A''$ from $A'$. By similar triangle $AIC' \sim A'A''C$, you have

$$AI \cdot A'C =2Rr$$

N. S.
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