Let A B and A C be drawn; as also any incident line D C, with its reflected line C F; and let the angle E C D be made double to the angle A; and let H B be drawn parallel to E C, and produced till it meet with D C produced in I. Lastly, producing A B indefinitely to K, let G B be drawn so that the angle G B K may be equal to the angle H B K, and then G B will be the reflected line of the incident line H B. I say, D C and H B are two incident lines, whose reflected lines C F and B G are parallel. For seeing the angle E C D is double to the angle B A C, the angle H I C is also, by reason of the parallels E C and H I, double to the same B A C; therefore also F C and G B, namely, the lines reflected from the incident lines D C and H B, are parallel. Wherefore the first thing required is done. Secondly, let it be required to draw to the points B and C two strait lines of incidence, so that the lines reflected from them may contain the given angle Z. To the angle E C D made at the point C, let there be added on one side the angle D C L equal to half Z, and on the other side the angle E C M equal to the angle D C L; and let the strait line B N be drawn parallel to the strait line C M; and let the angle K B O be made equal to the angle N B K; which being done, B O will be the line of reflection from the line of incidence N B. Lastly, from the incident line L C, let the reflected line C O be drawn, cutting B O at O, and making the angle C O B. I say, the angle C O B is equal to the angle Z. Let N B be produced till it meet with the strait line L C produced in P. Seeing, therefore, the angle L C M is, by construction, equal to twice the angle B A C, together with the angle Z; the angle N P L, which is equal to L C M by reason of the parallels N P and M C, will also be equal to twice the same angle B A C, together with the angle Z. And seeing the two strait lines O C and O B fall from the point O upon the points C and B; and their reflected lines L C and N B meet in the point P; the angle N P L will be equal to twice the angle B A C together with the angle C O B. But I have already proved the angle N P L to be equal to twice the angle B A C together with the angle Z. Therefore the angle C O B is equal to the angle Z; wherefore, two points in the circumference of a circle being given, I have drawn, &c.; which was to be done. But if it be required to draw the incident lines from a point within the circle, so that the lines reflected from them may contain an angle equal to the angle Z, the same method is to be used, saving that in this case the angle Z is not to be added to twice the angle B A C, but to be taken from it.