TITLE: Proving $R\ge 2 r$ using synthetic geometry
QUESTION [7 upvotes]: 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.

REPLY [8 votes]: 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
$$AI \cdot A'C =2Rr$$