Which is demonstrated thus.
[Illustration] Let CO be perpendicular to CR, and across the angle OCG let OK be adjusted, equal to N and perpendicular to CO, and let there be drawn the straight line KI, which if it is demonstrated to be a tangent to the Ellipse at I, it will be evident by the things heretofore explained that CI is the refraction of the ray RC.

Now since the angle RCO is a right angle, it is easy to see that the right-angled triangles RCV, KCO, are similar.

As then, CK is to KO, so also is RC to CV.

But KO is equal to N, and RC to CG: then as CK is to N so will CG be to CV.

But as N is to CG, so, by construction, is CV to CD.