propagation of the wave AC at the moment when the piece C of it has
arrived at B. For there is no other line which like BN is a common
tangent to all the aforesaid circles, except BG below the plane AB;
which line BG would be the propagation of the wave if the movement
could have spread in a medium homogeneous with that which is above the
plane. And if one wishes to see how the wave AC has come successively
to BN, one has only to draw in the same figure the straight lines KO
parallel to BN, and the straight lines KL parallel to AC. Thus one
will see that the straight wave AC has become broken up into all the
OKL parts successively, and that it has become straight again at NB.

Now it is apparent here that the angle of reflexion is made equal to
the angle of incidence. For the triangles ACB, BNA being rectangular
and having the side AB common, and the side CB equal to NA, it follows
that the angles opposite to these sides will be equal, and therefore
also the angles CBA, NAB. But as CB, perpendicular to CA, marks the
direction of the incident ray, so AN, perpendicular to the wave BN,
marks the direction of the reflected ray; hence these rays are equally
inclined to the plane AB.

But in considering the preceding demonstration, one might aver that it
is indeed true that BN is the common tangent of the circular waves in
the plane of this figure, but that these waves, being in truth
spherical, have still an infinitude of similar tangents, namely all
the straight lines which are drawn from the point B in the surface
generated by the straight line BN about the axis BA. It remains,
therefore, to demonstrate that there is no difficulty herein: and by
the same argument one will see why the incident ray and the reflected
ray are always in one and the same plane perpendicular to the
reflecting plane. I say then that the wave AC, being regarded only as