Suppose is the curvature line, denote the tangent vector of , and let B be the unit tangent vector of M

that is orthogonal to T. Thus T, B is the two principal directions forming a basis of the tangent plane, and denote K, k the corresponding principal curvatures.

In this basis the second fundamental form of M is a diagonal matrix

For the surface , the two coordinate tangent vectors are:

and

And since

From the above 3 equations we get (this can also be obtained by the Frenet Serret formulas)

So and

Then the normal vector of this surface is orthogonal to both T and U thus is constant along the v-lines on the surface, which implies the differential of the Gauss map is degenerate along U, thus having a 0 Gaussian curvature and hence developable.

I'll try to show the converse in a new post.