= Solution
\b[The asserted necessity of equal masses is false.] Equal unsigned outgoing angles imply equal longitudinal <momenta> when the transverse <momenta> cancel; they do not imply equal final speeds or <energies> for unequal masses.
An explicit counterexample uses $c=1$, masses $M_1=m$ and $M_2=2m$, initial projectile <energy> $E_1=2m$, and a stationary target. The initial total <energy> is $4m$ and <momentum> is $(\sqrt3m,0,0)$. Choose final <four-momenta>
$$
\boxed{P_1'=m\left(\frac{13}8,\frac{\sqrt3}2,\frac{\sqrt{57}}8,0\right),\qquad
P_2'=m\left(\frac{19}8,\frac{\sqrt3}2,-\frac{\sqrt{57}}8,0\right).}
$$
They sum to the incoming <four-momentum>. Their squared norms are respectively $m^2$ and $4m^2$, so the collision is elastic. Both trajectories make the same acute angle with positive $x$:
$$
\boxed{\tan\theta_1=\tan\theta_2=\frac{\sqrt{19}}4,\qquad M_1\ne M_2.}
$$
Thus no proof of the printed claim is possible.
The general <centre-of-mass frame> construction shows why. Let $p^*$ be the outgoing <momentum> magnitude there, $\theta^*$ the projectile scattering angle, and $E_1^*,E_2^*$ the two <energies>. After a boost $\beta_{\mathrm{CM}}$ along $x$, the equal-angle condition for two forward nonzero trajectories is
$$
p^*\cos\theta^*+\beta_{\mathrm{CM}}E_1^*
=-p^*\cos\theta^*+\beta_{\mathrm{CM}}E_2^*.
$$
With $E_1$ the initial projectile <energy>, its solution is
$$
\boxed{\cos\theta^*=
\frac{M_2^2-M_1^2}{2M_2(E_1+M_2)}\quad(c=1),}
$$
provided the right side lies between minus and plus one and both final longitudinal <momenta> are positive. Equal masses are one allowed case, giving $\theta^*=\pi/2$; they are not necessary. For the counterexample the right side is $3/16$. Even equal masses do not force equal angles for every scattering direction: the symmetric outgoing configuration is an additional choice.
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