ILL 1990 Problem 82

In a triangle, a symmedian is a line through a vertex that is symmetric to the median with the respect to the internal bisector (all relative to the same vertex). In the triangle $ABC$, the median $m_a$ meets $BC$ at $A'$ and the circumcircle again at $A_1$. The symmedian $s_a$ meets $BC$ at $M$ and the circumcircle again at $A_2$. Given that the line $A_1A_2$ contains the circumcenter $O$ of the triangle, prove that:

(a) $\frac{AA'}{AM} = \frac{b^2+c^2}{2bc} ;$

(b) $1+4b^2c^2 = a^2(b^2+c^2)$