Is the hyperbolic or spherical analogy of the following Euclidean fact, true?

Two triangles with equal corresponding medians are congruent.

More precisely: Assume that $\Delta ABC$ and $ \Delta A'B'C'$ are two triangles in the hyperbolic space $\mathbb{H}^2$ or elliptic space $\mathbb{S}^2$ such that $$AM_1=A'M_1',\; BM_2=B'M_2',\;CM_3=C'M_3'$$ where $M_i (M_i')$ in $XM_i(X'M_i')$ is the mid point of the edge opposite to the vertex $X(X')$, respectively.

Under this condition, is there an isometry of the corresponding space carring the first triangle to the second one?

Is the hyperbolic or spherical analogy of the following Euclidean fact, true?

Two triangles with equal corresponding medians are congruent.

More precisely: Assume that $\Delta ABC$ and $ \Delta A'B'C'$ are two triangles in the hyperbolic space $\mathbb{H}^2$ or elliptic space $\mathbb{S}^2$ such that $$AM_1=A'M_1',\; BM_2=B'M_2',\;CM_3=C'M_3'$$ where $M_i (M_i')$ in $XM_i(X'M_i')$ is the mid point of the edge opposite to the vertex $X(X')$, respectively.

Under this condition, is there an isometry of the corresponding space carrying the first triangle to the second one?

Is the hyperbolic or spherical analogy of the following Euclidean fact, true?

Two triangles with equal corresponding medians are congruent.

More precisely: Assume that $\Delta ABC$ and $ \Delta A'B'C'$ are two triangles in the hyperbolic space $\mathbb{H}^2$ or elliptic space $\mathbb{S}^2$ such that $$AM_1=A'M_1',\; BM_2=B'M_2',\;CM_3=C'M_3'$$ where $M_i (M_i')$ in $XM_i(X'M_i')$ is the mid point of the opposite edge to the vertex $X(X')$, respectively.

Under this condition, is there an isometry of the corresponding space carring the first triangle to the second one?

Is the hyperbolic or spherical analogy of the following Euclidean fact, true?

Two triangles with equal corresponding medians are congruent.

More precisely: Assume that $\Delta ABC$ and $ \Delta A'B'C'$ are two triangles in the hyperbolic space $\mathbb{H}^2$ or elliptic space $\mathbb{S}^2$ such that $$AM_1=A'M_1',\; BM_2=B'M_2',\;CM_3=C'M_3'$$ where $M_i (M_i')$ in $XM_i(X'M_i')$ is the mid point of the edge opposite to the vertex $X(X')$, respectively.

Under this condition, is there an isometry of the corresponding space carring the first triangle to the second one?