= Solution
Work in the parent's <rest frame>, with $M_A>0$. <Four-momentum conservation> gives
$$
E_B+E_C=M_Ac^2,\qquad \mathbf p_B=-\mathbf p_C,
$$
so both daughter <momenta> have magnitude $p$. Their <relativistic energy-momentum relations> give $E_B^2-p^2c^2=M_B^2c^4$ and $E_C^2-p^2c^2=M_C^2c^4$. Subtracting these two relations and using the total <energy> yields
$$
\boxed{E_B=\frac{M_A^2+M_B^2-M_C^2}{2M_A}c^2,\qquad
E_C=\frac{M_A^2+M_C^2-M_B^2}{2M_A}c^2.}
$$
Substitution back gives the <relativistic two-body decay> momentum:
$$
\boxed{p=\frac{c}{2M_A}\sqrt{\bigl[M_A^2-(M_B+M_C)^2\bigr]\bigl[M_A^2-(M_B-M_C)^2\bigr]},\qquad
\mathbf p_B=p\mathbf n,\quad\mathbf p_C=-p\mathbf n.}
$$
The unit <vector> $\mathbf n$ is not fixed by <four-momentum conservation> alone. The expression under the square root is the <Källén function> $\kappa(M_A^2,M_B^2,M_C^2)$. The physical <two-body decay threshold> is $M_A\geq M_B+M_C$; for massive daughters equality gives zero outgoing <momentum>. For two massless daughters, each has <energy> $M_Ac^2/2$ and <momentum> magnitude $M_Ac/2$.
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