Let $f$ be an isometry (i.e a diffeomorphism which preserves the Riemannian metrics) between Riemannian manifolds $(M,g)$ and $(N,h).$
One can argue that $f$ also preserves the induced metrics $d_1, d_2$ on $M, N$ from $g, h$ resp. that is, $d_1(x,y)=d_2(f(x),f(y))$ for $x,y \in M.$ Then, it's easy to show that $f$ sends geodesics on $M$ to geodesics on $N,$ using the length minimizing property of geodesics and that $f$ is distance-preserving.
My question,
Is it possible to derive the result without using the distance-preserving property of isometries, by merely the definition?
What I have found so far;
Let $\gamma : I \to M$ be a geodesic on $M$, i.e. $\frac{D}{dt}(\frac{d\gamma}{dt})=0,$ where $\frac{D}{dt}$ is the covariant derivative and $ \frac{d\gamma}{dt}:=d\gamma(\frac{d}{dt}).$ Let $t_0 \in I,$ we have to show that $\frac{D}{dt}(\frac{d(f \circ \gamma)}{dt})=0$ at $t=t_0,$ or $\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}))=0.$
We also know that
$$ \langle \frac{d \gamma}{dt}|_{t=t_0},\frac{d \gamma}{dt}|_{t=t_0}\rangle_{\gamma(t_0)}= \langle df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})\rangle_{f(\gamma(t_0))}.$$
Since $\frac{d}{dt} \langle \frac{d \gamma}{dt}|_{t=t_0},\frac{d \gamma}{dt}|_{t=t_0}\rangle_{\gamma(t_0)}=2 \langle \frac{D}{dt}(\frac{d \gamma}{dt}|_{t=t_0}),\frac{d \gamma}{dt}|_{t=t_0}\rangle_{\gamma(t_0)}=0,$ therefore
$$\frac{d}{dt} \langle df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})\rangle_{f(\gamma(t_0))}=$$
$$2\langle \frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})\rangle _{f(\gamma(t_0))}=0.$$
How can I conclude from $\langle\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0})), df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}) \rangle _{f(\gamma(t_0))}=0$ that $\frac{D}{dt}(df_{\gamma(t_0)}(\frac{d \gamma}{dt}|_{t=t_0}))=0?$