Johan Gunardi wrote: Given triangle $ ABC$. Let the tangent lines of the circumcircle of $ AB$ at $ B$ and $ C$ meet at $ A_0$. Define $ B_0$ and $ C_0$ similarly. a) Prove that $ AA_0,BB_0,CC_0$ are concurrent. b) Let $ K$ be the point of concurrency. Prove that $ KG\parallel BC$ if and only if $ 2a^2 = b^2 + c^2$.

Hope that point $ b)$ will be clear as soon as possible(may be next year=)). By th way Happy New Year everyone!!!!!!

a) clearly the concur in the gergonne point of the tangential triangle. It's also well-know that $ AA_0$ is a symmedian so K is also the symmedian point of ABC. b) Let M,N,L the midpoint of BC,CA,AB. Let J the midpoint of the altitude from A, so is know that K stay on JM i.e. in the medial triangle MNL, K is the isotomic conjugate of the orthocenter of MNL. But the centroid G of ABC is also the centroid of MNL so the thesis becomes: Let Q the isotomic cojugate of ABC and let AQ,BQ,CQ meet BC,CA,AB in D,E,F. Prove that $ \frac{AQ}{QD} = 2 \ \Longleftrightarrow \ 2a^2 = b^2 + c^2$. for menelaus on BDA and trasversal FQ we have: $ \frac{AQ}{QD} = \frac{BC}{CD} \cdot \frac{AF}{FB} = \frac{a}{c \cos B} \cdot \frac{a \cos B}{b \cos A} = \frac{a^2}{bc \cos A}$ so $ \frac{AQ}{QD} = 2 \ \Longleftrightarrow \ a^2 = 2 bc \cos A \ \Longleftrightarrow \ a^2 = b^2 + c^2 -a^2 \ \Longleftrightarrow \ 2a^2 = b^2 + c^2$