Problem 6

Let the line $AA_1$ meet $\omega_b$ again at $X$ and $\omega_c$ again at $Y$, and use directed ratios $k_a=\dfrac{AX}{AY}$. The circles $\omega_b,\omega_c$ are distinct and share $A_1,A_2$, so they are not concentric. The coaxiality lemma says that the locus $L_a:\operatorname{Pow}(P,\omega_b)=k_a\operatorname{Pow}(P,\omega_c)$ is a circle when $k_a\ne1$, and is a line when $k_a=1$. Here $A_1,A_2\in L_a$ because both powers are zero, while on $AA_1$ the secant-power theorem gives $\operatorname{Pow}(A,\omega_b)=AX\cdot AA_1$ and $\operatorname{Pow}(A,\omega_c)=AY\cdot AA_1$, so $A\in L_a$ by the definition of $k_a$. The points $A,A_1,A_2$ are noncollinear, hence $L_a$ cannot be a line; thus $k_a\ne1$ and $L_a$ is exactly $\mathcal C_a=(AA_1A_2)$. The directed-angle chase in the two auxiliary circles gives $\angle XAB_1=30^\circ-\beta$, $\angle A_1XB_1=30^\circ+\gamma$, and, similarly, $\angle YAC_1=30^\circ-\gamma$, $\angle A_1YC_1=30^\circ+\beta$. Applying the sine rule in $\triangle AB_1X$ and $\triangle AC_1Y$ therefore yields $\dfrac{AX}{AB_1}=\dfrac{\sin(\beta+\gamma)}{\sin(30^\circ+\gamma)}$ and $\dfrac{AY}{AC_1}=\dfrac{\sin(\beta+\gamma)}{\sin(30^\circ+\beta)}$. These are the required directed-angle calculations, not an appeal to an unlabeled diagram. Define $\mathcal C_b,\mathcal C_c$ and $k_b,k_c$ cyclically.