CHAPTER XIX.. Of Angles of Incidence and Reflection,

OF ANGLES OF INCIDENCE AND REFLECTION, EQUAL BY SUPPOSITION.

1. If two strait lines falling upon another strait line be parallel, the lines reflected from them shall also be parallel.—2. If two strait lines drawn from one point fall upon another strait line, the lines reflected from them, if they be drawn out the other way, will meet in an angle equal to the angle made by the lines of incidence.—3. If two strait parallel lines, drawn not oppositely, but from the same parts, fall upon the circumference of a circle, the lines reflected from them, if produced they meet within the circle, will make an angle double to that which is made by two strait lines drawn from the centre to the points of incidence.—4. If two strait lines drawn from the same point without a circle fall upon the circumference, and the lines reflected from them being produced meet within the circle, they will make an angle equal to twice that angle, which is made by two strait lines drawn from the centre to the points of incidence, together with the angle which the incident lines themselves make.—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.—6. If through any one point two unequal chords be drawn cutting one another, and the centre of the circle be not placed between them, and the lines reflected from them concur wheresoever, there cannot through the point, through which the two former lines were drawn, be drawn any other strait line whose reflected line shall pass through the common point of the two former lines reflected.—7. In equal chords the same is not true.—8. Two points being given in the circumference of a circle, to draw two strait lines to them, so that their reflected lines may contain any angle given.—9. If a strait line falling upon the circumference of a circle be produced till it reach the semidiameter, and that part of it, which is intercepted between the circumference and the semidiameter, be equal to that part of the semidiameter which is between the point of concourse and the centre, the reflected line will be parallel to the semidiameter.—10. If from a point within a circle, two strait lines be drawn to the circumference, and their reflected lines meet in the circumference of the same circle, the angle made by the reflected lines will be a third part of the angle made by the incident lines.

Whether a body falling upon the superficies of another body and being reflected from it, do make equal angles at that superficies, it belongs not to this place to dispute, being a knowledge which depends upon the natural causes of reflection; of which hitherto nothing has been said, but shall be spoken of hereafter.

In this place, therefore, let it be supposed that the angle of incidence is equal to the angle of reflection; that our present search may be applied, not to the finding out of the causes, but some consequences of the same.

I call an angle of incidence, that which is made between a strait line and another line, strait or crooked, upon which it falls, and which I call the line reflecting; and an angle of reflection equal to it, that which is made at the same point between the strait line which is reflected and the line reflecting.

1. If two strait lines, which fall upon another strait line, be parallel, their reflected lines shall be also parallel.

Let the two strait lines A B and C D (in fig. 1), which fall upon the strait line E F, at the points B and D, be parallel; and let the lines reflected from them be B G and D H. I say, B G and D H are also parallel.

For the angles A B E and C D E are equal by reason of the parallelism of A B and C D; and the angles G B F and H D F are equal to them by supposition; for the lines B G and D H are reflected from the lines A B and C D. Wherefore B G and D H are parallel.

2. If two strait lines drawn from the same point fall upon another strait line, the lines reflected from them, if they be drawn out the other way, will meet in an angle equal to the angle of the incident lines.

From the point A (in fig. 2) let the two strait lines A B and A D be drawn; and let them fall upon the strait line E K at the points B and D; and let the lines B I and D G be reflected from them. I say, I B and G D do converge, and that if they be produced on the other side of the line E K, they shall meet, as in F; and that the angle B F D shall be equal to the angle B A D.

For the angle of reflection I B K is equal to the angle of incidence A B E; and to the angle I B K its vertical angle E B F is equal; and therefore the angle A B E is equal to the angle E B F. Again, the angle A D E is equal to the angle of reflection G D K, that is, to its vertical angle E D F; and therefore the two angles A B D and A D B of the triangle A B D are one by one equal to the two angles F B D and F D B of the triangle F B D; wherefore also the third angle B A D is equal to the third angle B F D; which was to be proved.

Coroll. I. If the strait line A F be drawn, it will be perpendicular to the strait line E K. For both the angles at E will be equal, by reason of the equality of the two angles A B E and F B E, and of the two sides A B and F B.

Coroll. II. If upon any point between B and D there fall a strait line, as A C, whose reflected line is C H, this also produced beyond C, will fall upon F; which is evident by the demonstration above.

3. If from two points taken without a circle, two strait parallel lines, drawn not oppositely, but from the same parts, fall upon the circumference; the lines reflected from them, if produced they meet within the circle, will make an angle double to that which is made by two strait lines drawn from the centre to the points of incidence.

Let the two strait parallels A B and D C (in fig. 3) fall upon the circumference B C at the points B and C; and let the centre of the circle be E; and let A B reflected be B F, and D C reflected be C G; and let the lines F B and G C produced meet within the circle in H; and let E B and E C be connected. I say the angle F H G is double to the angle B E C.

For seeing A B and D C are parallels, and E B cuts A B in B, the same E B produced will cut D C somewhere; let it cut it in D; and let D C be produced howsoever to I, and let the intersection of D C and B F be at K. The angle therefore I C H, being external to the triangle C K H, will be equal to the two opposite angles C K H and C H K. Again, I C E being external to the triangle C D E, is equal to the two angles at D and E. Wherefore the angle I C H, being double to the angle I C E, is equal to the angles at D and E twice taken; and therefore the two angles C K H and C H K are equal to the two angles at D and E twice taken. But the angle C K H is equal to the angles D and A B D, that is, D twice taken; for A B and D C being parallels, the altern angles D and A B D are equal. Wherefore C H K, that is the angle F H G is also equal to the angle at E twice taken; which was to be proved.

Coroll. If from two points taken within a circle two strait parallels fall upon the circumference, the lines reflected from them shall meet in an angle, double to that which is made by two strait lines drawn from the centre to the points of incidence. For the parallels A B and I C falling upon the points B and C, are reflected in the lines B H and C H, and make the angle at H double to the angle at E, as was but now demonstrated.

4. If two strait lines drawn from the same point without a circle fall upon the circumference, and the lines reflected from them being produced meet within the circle, they will make an angle equal to twice that angle, which is made by two strait lines drawn from the centre to the points of incidence, together with the angle which the incident lines themselves make.

Let the two strait lines A B and A C (in fig. 4) be drawn from the point A to the circumference of the circle, whose centre is D; and let the lines reflected from them be B E and C G, and, being produced, make within the circle the angle H; also let the two strait lines D B and D C be drawn from the centre D to the points of incidence B and C. I say, the angle H is equal to twice the angle at D together with the angle at A.

For let A C be produced howsoever to I. Therefore the angle I C H, which is external to the triangle C K H, will be equal to the two angles C K H and C H K. Again, the angle I C D, which is external to the triangle C L D, will be equal to the two angles C L D and C D L. But the angle I C H is double to the angle I C D, and is therefore equal to the angles C L D and C D L twice taken. Wherefore the angles C K H and C H K are equal to the angles C L D and C D L twice taken. But the angle C L D, being external to the triangle A L B, is equal to the two angles L A B and L B A; and consequently C L D twice taken is equal to L A B and L B A twice taken. Wherefore C K H and C H K are equal to the angle C D L together with L A B and L B A twice taken. Also the angle C K H is equal to the angle L A B once and A B K, that is, L B A twice taken. Wherefore the angle C H K is equal to the remaining angle C D L, that is, to the angle at D, twice taken, and the angle L A B, that is, the angle at A, once taken; which was to be proved.

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.

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.

Coroll. If the angle A be greater than twice the angle D, their reflected lines will diverge. For, by the corollary of the third proposition, if the angle A be equal to twice the angle D, the reflected lines B E and C E will be parallel; and if it be less, they will concur, as has now been demonstrated. And therefore, if it be greater, the reflected lines B E and C E will diverge, and consequently, if they be produced the other way, they will concur and make an angle equal to the excess of the angle A above twice the angle D; as is evident by art. 4.

6. If through any one point two unequal chords be drawn cutting one another, either within the circle, or, if they be produced, without it, and the centre of the circle be not placed between them, and the lines reflected from them concur wheresoever; there cannot, through the point through which the former lines were drawn, be drawn another strait line, whose reflected line shall pass through the point where the two former reflected lines concur.

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.

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.

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.

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.

9. If a strait line, falling upon the circumference of a circle, be produced till it reach the semidiameter, and that part of it which is intercepted between the circumference and the semidiameter be equal to that part of the semidiameter which is between the point of concourse and the centre, the reflected line will be parallel to the semidiameter.

Let any line A B (in the 9th figure) be the semidiameter of the circle whose centre is A; and upon the circumference B D let the strait line C D fall, and be produced till it cut A B in E, so that E D and E A may be equal; and from the incident line C D let the line D F be reflected. I say, A B and D F will be parallel.

Let A G be drawn through the point D. Seeing, therefore, E D and E A are equal, the angles E D A and E A D will also be equal. But the angles F D G and E D A are equal; for each of them is half the angle E D H or F D C. Wherefore the angles F D G and E A D are equal; and consequently D F and A B are parallel; which was to be proved.

Coroll. If E A be greater then E D, then D F and A B being produced will concur; but if E A be less than E D, then B A and D H being produced will concur.

10. If from a point within a circle two strait lines be drawn to the circumference, and their reflected lines meet in the circumference of the same circle, the angle made by the lines of reflection will be a third part of the angle made by the lines of incidence.

From the point B (in the 10th figure) taken within the circle whose centre is A, let the two strait lines B C and B D be drawn to the circumference; and let their reflected lines C E and D E meet in the circumference of the same circle at the point E. I say, the angle C E D will be a third part of the angle C B D.

Let A C and A D be drawn. Seeing, therefore, the angles C E D and C B D together taken are equal to twice the angle C A D (as has been demonstrated in the 5th article); and the angle C A D twice taken is quadruple to the angle C E D; the angles C E D and C B D together taken will also be equal to the angle C E D four times taken; and therefore if the angle C E D be taken away on both sides, there will remain the angle C B D on one side, equal to the angle C E D thrice taken on the other side; which was to be demonstrated.

Coroll. Therefore a point being given within a circle, there may be drawn two lines from it to the circumference, so as their reflected lines may meet in the circumference. For it is but trisecting the angle C B D, which how it may be done shall be shown in the following chapter.

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Vol. 1. Lat. & Eng. C. XIX. Fig. 1-10 ]

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