4

Given two disjointed circles $c_1$ and $c_2$, external to each other, let $A$ be the meeting of their internal tangents and let $K$ be the orthogonal projection of $A$ in one the common external tangents between $c_1$ and $c_2$. Prove that $AK$ bissects angle $O_1KO_2$.

enter image description here

I managed to solve it by realizing that $\frac{KT_1}{KT_2} = \frac{r_1}{r_2}$, which clearly solves the problem when one look at $\triangle O_1T_1K$ and $\triangle O_2KT_2$. But this problem seems to have a simpler solution. What do you think?

hellofriends
  • 2,048
  • 1
    why simpler solution? KO1/AO1 = KO2/AO2? – Moti Jan 12 '22 at 07:20
  • 1
    Similarity of triangles is a non-trivial concept; it lets conclude about angles from lengths and vice-versa. I don't think there would be a proof from simply angle-chasing, which is a trivial thing (not much interplay between lengths and angles). – MyMolecules Jan 12 '22 at 12:20

0 Answers0