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the other extremity; but (1. 7.) this is impossible : therefore, if the base BC coincide with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore also the angle BAC coincides with the angle EDF, and (I. ax. 8.) is equal to it. Also the triangle ABC coincides with the triangle DEF, and (I. ax. 8.) is equal to it. Therefore, if two triangles, &c.
Let the triangle ABC be inverted with respect to the base BC, and let BC be applied to EF, so that B may
fall on E; then will C coincide with F, because BC is equal to EF; and the triangle ABC will take the position GEF, B EG being the same as AB, and FG the same as CA. Join DG; and, because (hyp.) DE and EG are equal, the angles EDG, EGD are (I. 5.) equal. It would be shown in a similar manner, that the angles FDG, FGD are equal ; and, therefore, (I. ax. 2.) the angle EGF, that is, BAC, is equal to EDF. But (hyp.) the sides BA, AC are respectively equal to ED, DF, and it has now been shown that the contained angles are equal; therefore (I. 4.) the triangles are equal, and the remaining angles in the one are respectively equal to the remaining angles in the other, that is, ABC to DEF, and ACB to DFE.*
PROP. IX. PROB.
To bisect a given rectilineal angle, that is, to divide it into two equal angles.
Let BAC be the given angle; it is required to bisect it. Take any point D in AB, and from AC cut off (I. 3.) A E equal to AD; join DE, and upon it describe (I. 1.) an equilateral triangle DEF, on the side remote from A ;t then join AF: AF bisects the angle BAC.
Because AD is equal to AE, and AF is common to the two triangles DAF, EAF; the two sides DA,
* Should DG fall on DF and FG, or on DE and EG, the proof would be had by means of one isosceles triangle. But should the point F fall within the isosceles triangle DEG, or E within DFG, two isosceles triangles would still be requisite, and the equality of the vertical angles would be proved by means of the third axiom.
† The expression, “ on the side remote from A,” is added in this edition, because, if the equilateral triangle were described on the other side its vertex might fall on A, in which case the solution would fail, as no second point in the bisecting line would then be determined.
In the practical construction of this problem, arcs may be described from D
AF are equal to the two EA, AF, each to each ; and the base DF is equal (const. and I. def. 17.) to the base EF: therefore, the angle DAF is equal (1. 8.) to EAF; wherefore, the given angle BAC is bisected by the line AF: which was to be done. *
PROP. X. PROB. +
To bisect a given finite straight line. Let AB be the given line; it is required to bisect it.
Describe (I. 1.) upon it an equilateral triangle ABC, and bisect (I. 9.) the angle ACB by the straight line CD. AB is bisected in the point D.I
Because AC is equal to CB, and CD common to the two triangles ACD, BCD; the two sides AC, CD are equal to BC, CD, each to each ; and the angle ACD is a equal (const.) to the angle BCD; therefore the base AD is equal (Î. 4.) to the base DB, and the line AB is bisected in the point D: which was to be done.
PROP. XI. PROB. §
To draw a straight line at right angles to a given straight line, from a given point in the same.
Let AB be a given straight line, and C a point given in it;
and E as centres with any radius greater than the half of DE, and the line joining either of their points of intersection with A will bisect the given angle. It is plain, that if the radius be taken equal to AD or AE, A must be joined with the remote point of intersection, as A and the other will coincide.
If the angles BAF, CAF were bisected, the given angle would be divided into four equal parts ; the bisection of wbich would divide it into eight equal parts; and thus the division, by successive bisections, might be continued without limit.
* Otherwise. Let DAE (1st fig. to prop. 5.) be the given angle; in AD assume any points, B and F, and cut off AC and AG, respectively equal to AB and AF: join BG and CF, and the straight line joining their point of intersection with A bisects the angle. The proof, which is easy, is left to exercise the learner.
† This problem is a particulare case of the ninth proposition of the sixth book.
In practice, the construction is effected more easily by describing arcs on both sides of AB, from A as centre, and with any radius greater than the half of AB; and then, by describing arcs intersecting them, with an equal radius, from B as centre: the line joining the two points of intersection will bisect AB. The proof is easy.
$ This proposition and the following contain the only two distinct cases of drawing a perpendicular to a given straight line through a given point; the first, when the point is in the line, and the second, when it is without it.
it is required to draw a straight line from C at right angles to AB.
Take any point D in AC, and make (I. 3.) CE equal to CD; upon DE describe (I. 1.) the equilateral triangle DFE, and (I. post. 1.) join FC; FC is the required line. *
Because DC is equal to CE, and FC common to the two triangles DCF, ECF; the two sides DC, CF are equal to the two EC, CF, each to each ; and the base DF is equal (const. and I. def. 17.) to the base EF; therefore, the angle DCF is equal (I. 8.) to the angle ECF; and they are adjacent angles. But, when the adjacent angles which one straight line makes with another straight line, are equal to one another, each of them is called (I. def. 8.) a right angle; therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the given point C in the given straight line AB, FC has been drawn at right angles to AB: which was to be done.
PROP. XII. PROB. To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it.
Let AB be the given straight line, which may be produced any length both ways, and let C be a point without it. It is required to draw a straight line from C perpendicular to AB.
Take any point D upon the other side of AB, and from the centre C, at the distance CD, describe (I. post. 3.) the circle EDF meeting AB in E and F: bisect (I. 10.) EF in G, and join CG; the straight line CG is the perpendicular required. +
Join CE, CF. Then because EG is equal to GF, and CG common to the two triangles EGC, FGC, the two sides EG, GC, are equal to the two FG, GC, each to each ; and the base CE is equal
In the practical construction of this problem, it is sufficient to describe arcs from D and E as centres, with any radius greater than DC, and to join their point of intersection with C. The proof follows from the construction and the eighth proposition. It will be a check on the manual operation, if intersections be made on both sides of A B, as these intersections and C should all be in the same straight line, as will appear from the corollary to the 14th proposition of this book.
+ In practice the construction will be rather more simple, by describing from F and E, when found, arcs on the remote side of A B from C, with any radius greater than the balf of FE, and joining their point of intersection with C.
The 12th proposition would be better placed after the 16th; as we cannot prove without the assistance of the 16th that there can be but one perpendicular drawn to a straight line from a point without it. We might then in fer also, that a circle can cut a straight line in only two points, since if it cut it in more than two, there might be more than one perpendicular.
(I. def. 30.) to the base CF, therefore the angle CGE is equal (I. 8.) to the angle CGF, and they are adjacent angles: therefore CG is perpendicular (I. def. 8.) to AB. Hence from the given point C a perpendicular CG has been drawn to the given line AB: which was to be done.
PROP. XIII. THEOR. The angles which one straight line makes with another upon the one side of it, are together equal to two right angles.
Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD; these are together equal to two right angles.
For, if the angle CBA be equal to ABD, each of them (I. def. 8.) is a right angle. But if not, from the point B draw (1. 11.) BE at right angles to CD ; * therefore the angles CBE, EBD are two right angles. Then, because DBE is equal to the two angles DBA, ABE together, add EBC to each of these equals ; therefore the angles CBE, EBD are equal (1. ax. 2.) to the three angles DBA, ABE, EBC. Again, because the angle CBA is equal to the two angles CBE, EBA, add ABD to these equals; therefore the angles CBA, ABD are equal to the three angles CBE, EBA, ABD; but CBE, EBD have been demonstrated to be equal to the same threo angles ; therefore (1. ax. 1.) the angles CBE, EBD are equal to the angles CBA, ABD; but CBE, EBD are two right angles; therefore CBA, ABD are together equal to two right angles. Wherefore the angles which one straight line, &c.
Cor. 1. From this it is manifest, that, if two straight lines cut one another, the four angles which they make at their point of intersection, are together equal to four right angles.
Cor. 2. And consequently that all the angles made by any number of lines meeting in one point, are together equal to four right angles.
PROP. XIV. THEOR. If, at a point in a straight line, two other straight lines on the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines are in one and the same straight line.
At the point B in the straight line AB, let the two straight * The substance of the remainder of this demonstration is, that the angles ABC, ABD are together equal to EBC, EBD, each pair being equal to the three angles CBE, EBA, ABD.
lines BC, BD, upon the opposite sides of AB, make the adjacent angles ABC, ABD equal together to two right angles : BD is in the same straight line with CB.
For if BD be not in the same straight line with CB, let BE be in the same straight line with it. Therefore, because AB makes angles with the straight line CBE, upon one side of it, those angles ABC, ABE, are together equal (1. 13.) to two right angles ; but the angles ABC, ABD are likewise together equal (hyp.) to two right angles: therefore (I. ax. 11.
and 1.) the angles CBA, ABE are equal to the angles CBA, ABD: take away the common angle ABC, and the remaining angle ABE is equal (I. ax. 3.) to the remaining angle ABD, a part to the whole, which is absurd. Therefore BĚ is not in the same straight line with BC: and in like manner it may be demonstrated, that no other can be in the same straight line with it but BD, which therefore is in the same straight line with CB. Wherefore, if at a point, &c.
Cor. If, at a point in a straight line, two other straight lines meet on the opposite sides of it, and make equal angles with the parts of it on opposite sides of the point, the two straight lines are in one and the same straight line.
For, let AEB, (fig. to next proposition) be a straight line, and let the angles AEC, BED be equal ; CE, ED are in the same straight line. For, by adding the angle CEB, to the equal angles AEC, BED, we have BED, BEC together equal to AEC, CEB, that is (I. 13.) to two right angles, and therefore, by this proposition, CE, ED are in the same straight line.
PROP. XV. THEOR.
If two straight lines cut one another, the vertical, or opposite, angles are equal.
Let the two straight lines AB, CD cut one another in E; the angle AEC is equal to the angle DEB, and CEB to AED.
Because the straight line AE, makes with CD the angles CEA, AED, these angles are together equal (I. 13.) to two right angles. Again, because DE makes with AB the angles AED, DEB, these also are together equal to two right angles ; and CEA, AED have been demonstrated to be equal to two right angles; wherefore (I. ax. 11, and 1.) the angles CEA, AED are equal to the angles AED, DEB. Take away the common angle AED, and (I. ax. 3.) the remaining angles CEA, DEB are equal. * In the same manner, it can be demonstrated that the angles CEB, AED are equal. Therefore, if two straight lines, &c.
* In the proof here given, the common angle is AED; and CEB, might with equal ease and propriety be made the common angle. In like manner, in prov