parmenides51 wrote:
Given triangle $ABC$ and points $P$. Let $A_1,B_1,C_1$ be the second points of intersection of straight lines $AP, BP, CP$ with the circumscribed circle of $ABC$. Let points $A_2, B_2, C_2$ be symmetric to $A_1,B_1,C_1$ wrt $BC,CA,AB$, respectively. Prove that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ are similar.
(A. Zaslavsky)
I don't see the proof of the problem in that blogspost so I post own .
$\textbf{Key Claim :}$ $\triangle PA_{2}B_{2}\sim{\triangle PA_{1}B_{1}}$
$\textbf{Proof :}$ Notice that $\angle PBA_2=\angle A_{1}AC-\angle CAB_2=\angle PAB_2$ , and since
\[\frac{A_{2}B}{B_{2}A}=\frac{BA_{1}}{AB_{1}}=\frac{PB}{PA}\]we have $\triangle PBA_{2}\sim \triangle PAB_{2}$ , so the length ratios to prove the claim are satisfied and so is the angle condition $\angle A_{2}PB_{2}=\angle APB=\angle A_{1}PB_{1}$ , proving the claim . $\square$
Analogously $\triangle PB_{2}C_{2}\sim \triangle PB_{1}C_{1}$ and $\triangle PC_{2}A_{2}\sim \triangle PC_{1}A_{1}$ , and hence $\triangle A_{1}B_{1}C_{1}\sim \triangle A_{2}B_{2}C_{2}$ , as desired . $\textbf{Q.E.D.}$