to each, and the bafe DF is equal to the base EF: therefore the angle D C F is equal [by prop. 8.] to the angle ECF; and thefe are adjacent angles. But when a right line standing upon a right line makes the adjacent angles equal to one an-A D C E B other, each of thefe equal angles is a right angle [by def. 1o.]. Therefore each of the angles DCF, FCE is a right angle. Therefore from a given point c, in a given line AB, the right line F c is drawn at right angles to the right line A B, PRO P. XII. PROB L. From a given point, without a given infinite right line, to draw a right line perpendicular to it. Let the given infinite right line be A B, and let the given point without the fame be c; it is required from the given point C without A B to draw a right line perpendicular to it. Let any point D be taken on the other fide of the line AB; and about the centre C, with the distance C D, let the circle E F G be described [by poft. 3.] and let the right line EG be bifected in the point H [by prop. 10.] and draw the right lines CG, CH, CE: I fay, that the right line C H is EB drawn from the given point C, without the given infinite right line A B perpendicular to it. For because G H is equal to the right line HE, and the right line HC is common; there are two right lines GH, HC, equal to two right lines E H, HC, each to each; and the bafe CG is equal to the bafe CE [by def. 15.] Therefore the angle C G H is equal to the angle EHC [by prop. 8.]; and these are adjacent angles. But when a right line ftanding upon a right line makes the adjacent angles equal to one another, each of the equal angles is a right angle, and the standing right line is called a perpendicular to that upon which it Itands. There Therefore upon the given infinite right line A B, from a given point c without it, is drawn the perpendicular C H to it. Which was to be done. It was very right in Euclid to fuppofe an infinite right line in this problem; for if it were a finite right line, a right line could not always be drawn from a given point out of it perpendicularly upon it. There are feveral practical ways of folving this problem, as well as the laft, but the most expeditious of all is by a fquare. PROP, XIII. THEOR. If a right line ftanding upon a right line makes angles, thefe angles fhall either be two right angles, or [both together] equal to two right angles &. For let any right line AB, ftanding upon the right line DC, make the angles C B A, A B D ; I fay, thefe angles will either be right angles, or [both together] equal to two right angles. For if the angle C B A be equal to the angle A B D; then will they be both right angles [by def. 10.] but if not, let the right line E B be drawn from the point B at right angles to the right line DC [by prop. II.] then the angles CBE, D B EBD are two right angles. And because the angle C BE is equal to both the angles C B A, AB E, if the common angle EBD be added, the angles CBE, BD are equal to the three angles CBA, A BE, EBD. Again, fince the angle DBA is equal to both the angles DBE, E B A, if the common angle A B C be added, the angles D B A, A BC will be equal to the three angles D B E, EBA, ABC. But the angles CBE, EBD have been proved to be equal to the fame three angles; and things which are equal to the fame thing, are equal to one another. Therefore the angles CBE, EBD are equal to the angles DBA, ABC; but the angles CBE, EBD are two right angles. Therefore the angles D BA, A B C are equal to two right angles. If therefore a right line ftanding upon a right line makes angles, thefe angles fhall either be two right angles, or [both together] equal to two right angles. Which was to be demonftrated. This propofition feems to depend upon a certain axiom or common notion, viz. that fo much as the angle A B D exceeds the right angle E B D, by fo much is the remaining angle A B C exceeded by the right angle E B C ; for as there the angle A B E is the excefs, fo is the fame angle A B E here the deficiency. Wherefore it neceffarily follows, that both the angles A B D, A B C are equal to two right angles, the one being just as much above a right angle, as the other wants of one. PROP. XIV. THEOR. If at a point in any right line two right lines drawn contrary ways do make, with the first line, the adjacent angles equal to two right angles, thofe right lines will be directly placed to one another, or both fall into one right linee. For at the point B in the right line A B let two right lines B C, B D drawn contrary ways, make, with AB, the adjoining angles A B C, A B D equal to two right angles: I fay, the right lines B D, C B will lie directly to one another, or both fall into one right line. For if B D lies not in the fame direction with B C, let [by poft. 2.] C B, B E both lie in the fame direction. Then because the right line A B stands upon the right A B D line C B E, the angles A B C, A B E are [by prop. 13.] equal to two, right angles; but [by the fuppofition] the angles A B C, A B D are E alfo equal to two right angles: therefore the angles CBA, ABE are equal to the angles C B A, Ꭺ Ᏼ Ꭰ. Take away the common angle A B C, then the remaining angle ABE will be equal to the remaining angle ABD, the lefs equal to the greater; which is impoffible: therefore, the right lines B E, BC are not both in one direction. is demonftrated after the fame manner, that no right line but B D can have the fame direction with the right line C B. Wherefore BC and B D lie both in the fame right line. It If therefore at a point in any right line, two right lines, drawn contrary ways, do make with the firft line the adjacent angles equal to two right angles, thofe right lines will be directly placed to one another, or both fall into one right line. Which was to be demonstrated. A B • This propofition would not generally have held true, if the two right lines, as R C, B D had been both on the fame fide the right line A B; for if the angle G B D, at the point в, made with the continuation B G of the right line AB, be equal to the angle A B C, made at the point в with the right line A B, the angles A B C, A B D taken together, are equal to two right angles. But the right lines B C, B D are far from coinciding; that is, both making one right line. Indeed, in this cafe, where the two right lines are both on one fide the first propofed right line, the propofition never can take place, unless the angles A B C, ABD be both right angles. PROP. XV. THEOR. G D If two right lines cut each other, they shall make the vertical angles equal to one another. For let two right lines A B, CD cut each other in the point E: I fay, the angle A EC is equal to the angle D E B, and the angle C E B equal to the A angle A E D. For because the right line A E ftands upon the right line c D, D making the angles C E A, A E D. Therefore the angles CE A, AED are [by prop. 13.] equal E B to two right angles. Again, because the right line DE ftands upon the right line A B, making the angles A E D, DE B, the angles A ED, D E B will be equal to two right angles. But the angles CEA, A E D have been proved to be equal to two right angles. Therefore the angles CEA, A E D are equal to the angles A E D, DE B. Take away the common angle A E D, then the remaining angle CEA is equal to the remaining angle B E D. In like manner it may be proved, that the angle C E B fhall be equal to the angle DE A. If therefore two right lines cut each other, they fhall make the vertical angles equal to one another. Which was to be demonstrated. Corol. From hence it is manifeft, that if never fo many right lines mutually cut one another in the fame point, they fhall make the angles at the point of interfection equal to four right angles. Any one fide of every triangle being produced, the outward angle is greater than either of the inward oppofite angles. Let the triangle be BA C, and let the fide BC be produced to D: I fay, the outward angle A C D is greater than either of the inward oppofite angles C B A, B A C. E F For bifect [by prop. 10. A c in the point E, and drawing the right line в E, produce it to F, and [by prop. 3:] put E F equal to B E, join F C, and continue out A c to G. Then because A E is equal to E C, and B E equal to the right line EF, there are two right lines A E, E B equal to two right lines c E, E F, each to each; the angle A E B allo is equal [by prop. 15.] to the angle FEC, for they are vertical angles. Therefore the bafe A B is equal to the bafe F C [by D prop. 4.] and the triangle A B E equal to the triangle F E C, and the remaining angles equal to the remaining angles each to each, which are oppofite to the equal fides therefore the angle B A E is equal to the angle E CF. But [by ax. 9.] the angle E C D greater than the angle ECF; therefore the angle A CD is greater than the angle B A E. In like manner, if B C be bifected, it will be demonftrated that the angle B C G, that is, [by prop. 15.] the angle A CD is greater than the angle A B C. ΒΕ is F G Therefore if any one fide of every triangle be produced, the |