Lemma: Given segments of lengths $x$, $y$, $z$, we can construct a segment with length $xy/z$. Proof: PoP. Thus we can construct $I_A$ because $AI \cdot AI_A = bc$. Then draw the circle with diameter $\overline{II_A}$, intersect it with the circle at $A$ with radius $b$ at $C_1, C_2$, and intersect it with the circle at $A$ with radius $c$ at $B_1$, $B_2$, so that $B_1$ and $C_1$ lie on the same side of $\overline{AI}$. Either $AB_1C_2$ or $AB_2C_1$ works.