= Solution
Let $\rho(x,y)$ be the common <boundary distance function>, with $x\ne y$. For each metric, the <simple Riemannian manifold> property gives a unique unit-speed <geodesic> from $x$ to $y$, smoothly depending on the endpoint pair. Denote its initial and terminal velocities by $v_i,w_i$.
The <first variation of geodesic energy>, or equivalently length variation at unit speed, gives the endpoint formula
$$
d_x\rho(\xi)=-g_i(v_i,\xi),\qquad
d_y\rho(\eta)=g_i(w_i,\eta)
\quad(\xi\in T_x\partial M,\ \eta\in T_y\partial M).
$$
For completeness, if $J$ is an endpoint variation along a geodesic, differentiating its length and integrating the <covariant derivative> by parts gives
$$
\delta L=\langle J,\dot\gamma\rangle\big|_0^L
-\int_0^L\langle J,D_t\dot\gamma\rangle\,dt.
$$
The integral vanishes by the <geodesic equation>, leaving precisely these two boundary terms and their opposite signs.
Because the full metrics agree on the boundary tangent spaces $T_xM$, the induced boundary metric $h$, unit spheres and inward unit normals agree there. The displayed covectors therefore determine the same tangential velocity components for the two metrics:
$$
(v_i)_T=-\operatorname{grad}_x^h\rho,\qquad
(w_i)_T=\operatorname{grad}_y^h\rho.
$$
Their normal components are fixed by unit length and the entry or exit sign. With inward unit normals $\nu_x,\nu_y$,
$$
v_i=(v_i)_T+\sqrt{1-| (v_i)_T|_h^2}\,\nu_x,
\qquad
w_i=(w_i)_T-\sqrt{1-| (w_i)_T|_h^2}\,\nu_y.
$$
Strict convexity makes these signs strict for distinct endpoints. Thus $v_1=v_2$ and $w_1=w_2$ for the same endpoint pair.
Now start with any strictly inward $(x,v)$ and let its $g_1$ exit point be $y$. The $g_2$ geodesic joining $x$ to this same $y$ has the same initial velocity $v$, so uniqueness of the <geodesic equation> identifies it with the $g_2$ trajectory launched from $(x,v)$. Its exit velocity is also the same. Hence
$$
\boxed{\alpha_{g_1}=\alpha_{g_2}.}
$$
This is the <boundary distance determines geodesic scattering> argument. The extension to outward vectors follows by the inverse scattering relation, and tangential vectors are fixed for both metrics. Equality of the travel times follows as well, since both equal $\rho(x,y)$.
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