Let any two unequal chords, as B K and C H (in [fig. 6]), be drawn through the point A in the circle B C; and let their reflected lines B D and C E meet in F; and let the centre not be between A B and A C; and from the point A let any other strait line, as A G, be drawn to the circumference between B and C. I say, G N, which passes through the point F, where the reflected lines B D and C E meet, will not be the reflected line of A G.
For let the arch B L be taken equal to the arch B G, and the strait line B M equal to the strait line B A; and L M being drawn, let it be produced to the circumference in O. Seeing therefore B A and B M are equal, and the arch B L equal to the arch B G, and the angle M B L equal to the angle A B G, A G and M L will also be equal, and, producing G A to the circumference in I, the whole lines L O and G I will in like manner be equal. But L O is greater than G F N, as shall presently be demonstrated; and therefore also G I is greater than G N. Wherefore the angles N G C and I G B are not equal. Wherefore the line G F N is not reflected from the line of incidence A G, and consequently no other strait line, besides A B and A C, which is drawn through the point A, and falls upon the circumference B C, can be reflected to the point F; which was to be demonstrated.
It remains that I prove L O to be greater than G N; which I shall do in this manner. L O and G N cut one another in P; and P L is greater than P G. Seeing now L P. P G :: P N. P O are proportionals, therefore the two extremes L P and P O together taken, that is L O, are greater than P G and P N together taken, that is, G N; which remained to be proved.
In equal chords the same is not true.
7. But if two equal chords be drawn through one point within a circle, and the lines reflected from them meet in another point, then another strait line may be drawn between them through the former point, whose reflected line shall pass through the latter point.
Let the two equal chords B C and E D (in the [7th figure]) cut one another in the point A within the circle B C D; and let their reflected lines C H and D I meet in the point F. Then dividing the arch C D equally in G, let the two chords G K and G L be drawn through the points A and F. I say, G L will be the line reflected from the chord K G. For the four chords B C, C H, E D and D I are by supposition all equal to one another; and therefore the arch B C H is equal to the arch E D I; as also the angle B C H to the angle E D I; and the angle A M C to its verticle angle F M D; and the strait line D M to the strait line G M; and, in like manner, the strait line A C to the strait line F D; and the chords C G and G D being drawn, will also be equal; and also the angles F D G and A C G, in the equal segments G D I and G C B. Wherefore the strait lines F G and A G are equal; and, therefore, the angle F G D is equal to the angle A G C, that is, the angle of incidence equal to the angle of reflection. Wherefore the line G L is reflected from the incident line C G; which was to be proved.
Coroll. By the very sight of the figure it is manifest, that if G be not the middle point between C and D, the reflected line G L will not pass through the point F.
Two points being given in the circumference of a circle, to draw two strait lines to them so as that their reflected lines may contain any angle given.
8. Two points in the circumference of a circle being given to draw two strait lines to them, so as that their reflected lines may be parallel, or contain any angle given.
In the circumference of the circle, whose centre is A, (in the [8th figure]) let the two points B and C be given; and let it be required to draw to them from two points taken without the circle two incident lines, so that their reflected lines may, first, be parallel.