Now we know that the piece of wave which is incident on G, advances thence along the line GD, since GV is the refraction of the ray LG.

Then during the time that this piece of wave has taken from G to D, the other piece which was at S has reached E, since GD, SE are equal.

But while the latter will advance from E to B, the piece of wave which was at D will have spread into the air its partial wave, the semi-diameter of which, DC (supposing this wave to cut the line DF at C), will be 3/2 of EB, since the velocity of light outside the medium is to that inside as 3 to 2.

Now it is easy to show that this wave will touch the arc BP at this point C.For since, by construction, FD + 3/2 DG + GL are equal to FB + 3/2 BA + AL; on deducting the equals LH, LA, there will remain FD + 3/2 DG + GH equal to FB + 3/2 BA.

And, again, deducting from one side GH, and from the other side 3/2 of AS, which are equal, there will remain FD with 3/2 DG equal to FB with 3/2 of BS.