For, considering AB as the radius of a circle, the Sine of the angle BAC is BC, and the Sine of the angle ABN is AN.

But the angle BAC is equal to DAE, since each of them added to CAE makes a right angle.

And the angle ABN is equal to NAF, since each of them with BAN makes a right angle.

Then also the Sine of the angle DAE is to the Sine of NAF as BC is to AN.

But the ratio of BC to AN was the same as that of the velocities of light in the substance which is towards AE and in that which is towards AF; therefore also the Sine of the angle DAE will be to the Sine of the angle NAF the same as the said velocities of light.
To see, consequently, what the refraction will be when the waves of light pass into a substance in which the movement travels more quickly than in that from which they emerge (let us again assume the ratio of 3 to 2), it is only necessary to repeat all the same construction and demonstration which we have just used, merely substituting everywhere 3/2 instead of 2/3.