Prove the Gaussian curvature $K=0$

Question

If two families of a geodesics on a surface intersect at a constant angle $\theta$, prove that the Gaussian curvature of the surface is zero, i.e. $K=0$.

Please explain how to show the question. Thank you for helping.

$K=k_1k_2$ for principal curvatures $k_1, k_2$.

Answer 1

HINT: Consider small parallelograms formed by these geodesics, and apply Gauss-Bonnet.

EDIT: Here's what I mean: Take two geodesics $\gamma_1$ and $\gamma_2$ from family A and two geodesics $\sigma_1$ and $\sigma_2$ from family B. Define points $P$, $Q$, $R$, and $S$ by taking $P=\gamma_1\cap\sigma_1$, $Q=\gamma_1\cap\sigma_2$, $R=\gamma_2\cap\sigma_2$, and $S=\gamma_2\cap\sigma_1$. (Draw a picture of all this.) Apply Gauss-Bonnet to this "parallelogram" formed by the pieces of the four geodesics joining $P$, $Q$, $R$, and $S$.