 Solution

Full equality of the Riemannian metrics at the boundary identifies their boundary unit bundles, outward normals and induced boundary Riemannian metrics. We show directly that the common boundary distance function determines both endpoint directions.

For distinct boundary points $x,y$, let $\gamma$ be the unique joining unit-speed geodesic, with initial velocity $v$ and final velocity $w$. The first variation of its length gives, for $a\in T_x\partial M$ and $b\in T_y\partial M$,Indeed, integration by parts in the length variation leaves just the endpoint terms, because the interior term contains ${\nabla }_{\dot{\gamma }}\dot{\gamma }=0$. The Riemannian metric is simple, so the off-diagonal endpoint dependence is smooth and these derivatives are valid.

Thus the tangential components are $v_T=-{\mathrm{grad}}_x^{\partial M}{d}_g(x,y)$ and $w_T={\mathrm{grad}}_y^{\partial M}{d}_g(x,y)$. Unit speed and the incoming/outgoing signs supply the missing normal components:All quantities on the right use only boundary distance and the common boundary Riemannian metric. Every strictly incoming vector for $g_1$ has some exit point $y$; the geodesic joining the same pair for $g_2$ has exactly the same reconstructed initial and final vectors. Hence both geodesic scattering relations give the same exit for that input. Tangential vectors are fixed, and the outgoing extensions agree as inverses. ThereforeIn fact their exit times also agree, since each is $d_g(x,y)$.