Coroll. If two strait converging lines, as I C and M B, fall upon the concave circumference of a circle, their reflected lines, as C H and B H, will meet in the angle H, equal to twice the angle D, together with the angle at A made by the incident lines produced. Or, if the incident lines be H B and I C, whose reflected lines C H and B M meet in the point N, the angle C N B will be equal to twice the angle D, together with the angle C K H made by the lines of incidence. For the angle C N B is equal to the angle H, that is, to twice the angle D, together with the two angles A, and N B H, that is, K B A. But the angles K B A and A are equal to the angle C K H. Wherefore the angle C N B is equal to twice the angle D, together with the angle C K H made by the lines of incidence I C and H B produced to K. [Sidenote: If two strait lines drawn from one point fall upon the concave circumference of a circle, and the angle they make be less than twice the angle at the centre, the lines reflected from them and meeting within the circle will make an angle, which being added to the angle of the incident lines will be equal to twice the angle at the centre.] 5. If two strait lines drawn from one point fall upon the concave circumference of a circle, and the angle they make be less than twice the angle at the centre, the lines reflected from them and meeting within the circle will make an angle, which being added to the angle of the incident lines, will be equal to twice the angle at the centre. Let the two lines A B and A C (in fig. 5), drawn from the point A, fall upon the concave circumference of the circle whose centre is D; and let their reflected lines B E and C E meet in the point E; also let the angle A be less than twice the angle D. I say, the angles A and E together taken are equal to twice the angle D. For let the strait lines A B and E C cut the strait lines D C and D B in the points G and H; and the angle B H C will be equal to the two angles E B H and E; also the same angle B H C will be equal to the two angles D and D C H; and in like manner the angle B G C will be equal to the two angles A C D and A, and the same angle B G C will be also equal to the two angles D B G and D. Wherefore the four angles E B H, E, A C D and A, are equal to the four angles D, D C H, D B G and D. If, therefore, equals be taken away on both sides, namely, on one side A C D and E B H, and on the other side D C H and D B G, (for the angle E B H is equal to the angle D B G, and the angle A C D equal to the angle D C H), the remainders on both sides will be equal, namely, on one side the angles A and E, and on the other the angle D twice taken. Wherefore the angles A and E are equal to twice the angle D.