How can I show that $\langle v , w \rangle = \lambda \langle d {\Phi}_p(v) , d {\Phi}_p(w) \rangle$ for any $\lambda \in \mathbb{R}$?

Question

Let $S$ and $S'$ be two surfaces and let $\Phi : S \to S'$ be a diffeomorphism. We suppose that $\Phi \circ \alpha$ is a geodesic in $S'$ for all geodesic $\alpha$ in $S$. If $S$ is connected and $\Phi$ is a conformal mapping, how can I show that there exists $\lambda \in \mathbb{R}$ such that $$ \langle v , w \rangle = \lambda \langle d {\Phi}_p(v) , d {\Phi}_p(w) \rangle \tag{1} $$ for all $p \in S$ and for each $v , w \in T_pS$? Since it is conformal mapping, then there exists $\lambda : S \to (0 , \infty)$ such that $$ \langle d {\Phi}_p(v) , d {\Phi}_p(w) \rangle = \lambda(p) \langle v , w \rangle\mbox{.} $$ Thus a way to show (1) is see that $\lambda$ is a constant function using that $S$ is connected and using that $\Phi$ transforms geodesic in $S$ to geodesics in $S'$. Is it possible to do this?

Answer 1

If $c,\ d$ are unit speed geodesic s.t. $c(0)=d(0)$ and $ c'(0)\perp d'(0)$, then $ \angle\ c(-t)d(t)c(t)$ goes to $\frac{\pi}{2}$.

If $C,\ D$ are images of $\Phi$, then assume that $|C'|=a,\ |D'|=b$. If $a=b$, then we are done. If not, consider $\angle\ C(-t)D(t) C(t)$ which does not go to $\frac{\pi}{2}$. It is a contradiction, since $\Phi$ is conformal.