= Solution
A <Lorentz invariant> has the same value when the same physical system is described in any inertial frame. The components of its <energy-momentum four-vector> change, but their <Minkowski norm> does not. In particular the <relativistic energy-momentum relation> is
$$
\boxed{E^2-c^2|\mathbf p|^2=M^2c^4,}
$$
where $M$ is the <rest mass>. Individual total <energy> and <momentum> are frame-dependent; the combination on the left is invariant.
For the fixed-target annihilation, the initial total laboratory <energy> is $E+Mc^2$ and its total <momentum> is the incident particle's $\mathbf p$, with $c^2p^2=E^2-M^2c^4$. The <centre-of-mass energy> available for new-particle creation is therefore
$$
\boxed{W=\sqrt{(E+Mc^2)^2-c^2p^2}
=\sqrt{2Mc^2(E+Mc^2)}.}
$$
In the <centre-of-mass frame> the total <momentum> vanishes, so $W$ can supply the new particles' total <rest energy> at threshold. One does not subtract the annihilated particles' original <rest energy>. The excess laboratory <energy> beyond $W$ is associated with motion of the whole final system in the <laboratory frame>.
For the subsequent <elastic collision>, the equal and opposite transverse components of final <momentum> imply equal final momentum magnitudes, and hence equal final <energies>, for the two identical particles. By <conservation of relativistic energy>, each has
$$
E_f=Mc^2+\frac T2.
$$
Their final momentum magnitude $p_f$ and the incident momentum magnitude $p_i$ obey
$$
c^2p_f^2=TMc^2+\frac{T^2}{4},\qquad
c^2p_i^2=T^2+2TMc^2.
$$
Longitudinal <momentum conservation> gives $p_i=2p_f\cos\theta$. It follows that the <equal-angle relativistic elastic scattering> relation is
$$
\boxed{\cos^2\theta=\frac{p_i^2}{4p_f^2}
=\frac{T+2Mc^2}{T+4Mc^2}.}
$$
This derivation assumes nonzero incident <kinetic energy> and the stated symmetric outgoing paths. The angle between those paths is $2\theta$, which is evaluated in the two following parts.
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