For in the circle A C (fig. 4) let the angle of inclination A B C be an angle of 45 degrees. Let C B be produced to the circumference in D; and let C E, the sine of the angle E B C, be drawn, to which let B F be taken equal in the separating line B G. B C E F will therefore be a parallelogram, and F E and B C, that is F E and B G equal. Let A G be drawn, namely the diagonal of the square whose side is B G, and it will be, as A G to E F so B G to B F; and so, by supposition, the density of the medium, in which C is, to the density of the medium in which D is; and so also the sine of the angle refracted to the sine of the angle of inclination. Drawing therefore F D, and from D the line D H perpendicular to A B produced, D H will be the sine of the angle of inclination. And seeing the sine of the angle refracted is to the sine of the angle of inclination, as the density of the medium, in which is C, is to the density of the medium in which is D, that is, by supposition, as A G is to F E, that is as B G is to D H; and seeing D H is the sine of the angle of inclination, B G will therefore be the sine of the angle refracted. Wherefore B G will be the refracted line, and lye in the plain separating superficies; which was to be demonstrated. Coroll. It is therefore manifest, that when the inclination is greater than 45 degrees, as also when it is less, provided the density be greater, it may happen that the refraction will not enter the thinner medium at all. [Sidenote: If a body be carried in a strait line upon another body, and do not penetrate it, but be reflected from it, the angle of reflection will be equal to the angle of incidence.] 8. If a body fall in a strait line upon another body, and do not penetrate it, but be reflected from it, the angle of reflection will be equal to the angle of incidence. Let there be a body at A (in fig. 5), which falling with strait motion in the line A C upon another body at C, passeth no further, but is reflected; and let the angle of incidence be any angle, as A C D. Let the strait line C E be drawn, making with D C produced the angle E C F equal to the angle A C D; and let A D be drawn perpendicular to the strait line D F. Also in the same strait line D F let C G be taken equal to C D; and let the perpendicular G E be raised, cutting C E in E. This being done, the triangles A C D and E C G will be equal and like. Let C H be drawn equal and parallel to the strait line A D; and let H C be produced indefinitely to I. Lastly let E A be drawn, which will pass through H, and be parallel and equal to G D. I say the motion from A to C, in the strait line of incidence A C, will be reflected in the strait line C E.