My kid was given this question. (Well, my statement of it actually includes some results that my kid had to find in previous parts of the question.)

Triangle $ABC$ is equilateral. $D$ is the middle of side $\overline{BC}$. $AD$ is the diameter of a circle centered at $O$. $\overline{AC}$ meets the circle at $F$. Prove that $AF=3CF$.

[I'd appreciate if someone could add a diagram. I can't at the moment.]

I can think of two ways to do this:

1. Draw $\overline{DF}$, prove it's an altitude in triangle $ADC$, and use similar triangles to find the sidelengths.
2. Use the fact that $DC^2=FC\cdot AC$.

Any other methods? I ask because my kid hasn't learned that latter theorem, or similar triangles.

posted almost 3 years ago

CC BY-NC-SA 4.0

msh210‭

292 reputation 9 4 43 4