BG-MO 1970 Problem 4

Let $\delta_0=\triangle A_0B_0C_0$ be a triangle. On each of the sides $B_0C_0$ , $C_0A_0$ , $A_0B_0$ , there are constructed squares in the halfplane, not containing the respective vertex $A_0,B_0,C_0$ and $A_1,B_1,C_1$ are the centers of the constructed squares. If we use the triangle $\delta_1=\triangle A_1B_1C_1$ in the same way we may construct the triangle $\delta_2=\triangle A_2B_2C_2$ ; from $\delta_2=\triangle A_2B_2C_2$ we may construct $\delta_3=\triangle A_3B_3C_3$ and etc. Prove that:

(a) segments $A_0A_1,B_0B_1,C_0C_1$ are respectively equal and perpendicular to $B_1C_1,C_1A_1,A_1B_1$ ; (b) vertices $A_1,B_1,C_1$ of the triangle $\delta_1$ lies respectively over the segments $A_0A_3,B_0B_3,C_0C_3$ (defined by the vertices of $\delta_0$ and $\delta_1$ ) and divide them in ratio $2:1$ .

K. Dochev