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point without it. It is required to draw a straight line perpendicular to AB from the point c.

Take any point d upon the other side of AB, from the cen

II tre c, at the distance cd, describe (Post. 3.) the circle EGF meeting AB in F, G; and bisect (1. 10.) FG in H, and join CF, CH, CG; the straight line ch, drawn from the given point c, is perpendicular to the given straight line AB.

Because Fh is equal to hd, and no common to the two triangles FHC, GHC, the two sides FH, HC are equal to the two GH, HC, each to each ; and the base CF is equal (1. Def. 15.) to the base cg; therefore the angle car is equal (1. 8.) to the angle CHG; and they are adjacent angles; but when a straight line standing on a straight line makes the adjacent angles equal to one another, each of them is a right angle, and the straight line which stands upon the other is called a perpendicular to it (Def. 10.); therefore from the given point c a perpendicular ch has been drawn to the given straight line AB. Which was to be done.

PROP. XIII. THEOR. The angles which one straight line makes with another

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the one side of it, are either two right angles, or are together equal to two right angles.

Let the straight line A B make with CD, upon one side of it, the angles CBA, ABD; these are either two right angles, or are together equal to two right angles.

For, if the angle cba be equal to ABD, each of them is a right (Def. 10.) angle; but, if not, from

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the point B draw BE at right angles (1. 11.) to cd; therefore the angles C BE, E B D, are two right angles (Def. 10.); D and because CBE is equal to the two angles CBA, ABE together, add the angle EBD to each of these equals; therefore the angles CBE, EBD are equal (Ax. 2.) to the three angles CBA, A BE, E B D. Again, because the angle DBA is equal to the two angles DB E, E B A, add to these equals the angle ABC; therefore the angles D B A, A B C are equal to the three angles D BE, E B A, A BC; but the angles C B E, E B D have been demonstrated to be equal to the same three angles ; and things that are equal to the same are equal (Ax. 1.) to one another; therefore the angles C B E, E B D are equal to the angles DBA, ABC; but C BE, EBD are two right angles ; therefore DBA, ABC are together equal to two right angles. Wherefore, when a straight line, &c. Q. E. D.

PROP. XIV. THEOR. If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.

At the point B in the straight line A B, let the two straight lines BC, BD upon

the opposite sides of A B, make the adjacent angles ABC, ABD,

E equal together to two right angles. BD is in the same

/B straight line with C B.

For, if BD be not in the same straight line with CB, let BE be in the same straight line with it;

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therefore, because the straight line AB makes agles with the straight lines C B E, upon one side of it, the angles A B C, A B E are together equal (1. 13.) to two right angles; but the angles A B C,

are likewise together equal to two right angles (Hyp.); therefore the angles CBA, A BE are equal to the angles CBA, A B D. Take away the common angle A B C, the remaining angle A B E is equal (Ax. 3.) to the remaining angle A B D, the less to the greater, which is impossible; therefore B E is not in the same straight line with B C. 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. Q. E. D.

PROP. XV. THEOR. If two straight lines cut one another, the vertical, or opposite, angles shall be equal.

Let the two straight lines A B, CD cut one another in the point E; the angle A E C shall be equal to the angle D E B, and C E B to EAD.

Because the straight line A E makes with co the angles C E A, A E D, these angles are together equal (1. 13.) to two right angles. Again, because the straight line C De makes with A B the angles A ED, DEB these also are together equal (1. 13.) to two right angles; and CEA, A ED have been demonstrated to be equal to two right angles ; wherefore the angles CEA, A E D are equal to the angles A ED, DE B. Take

away the common angle A E D, and the remaining angle CEA is equal (Ax. 3.) to the remaining angle DE B. In the same manner it can

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be demonstrated that the angles CEB, A ED are equal. Therefore if two straight lines, &c. Q. E. D.

Cor. 1. From this it is manifest, that, if two straight lines cut one another, the angles they make at the point where they cut, 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.

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PROP. XVI. THEOR. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

Let ABC be a triangle, and let its side Bc be produced to D, the exterior angle A c d is greater than either of the interior opposite angles CBA, BAC.

Bisect (1. 10.) AC in E, join B E and produce it to F, and make E F equal to BE; B join also FC, and produce AC to G.

Because A E is equal to EC and BE to EF; A E, E B are equal to CE, E F, each to each; and the angle A E B is equal (1. 15.) to the angle CEF, because they are opposite vertical angles; therefore the base A B is equal (1. 4.) to the base CF, and the triangle A E B to the triangle CF E, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite : wherefore the angle B A E is equal to the angle E CF; but the angle E cd is greater than the angle ECF; therefore the angle Acd is greater than B A E. In the same manner, if the side bc be

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bisected, it may be demonstrated that the angle BCG, that is (1. 15.), the angle ACD, is greater than the angle ABC. Therefore, if one side, &c.

Q. E. D. PROP. XVII. THEOR. Any two angles of a triangle are together less than two right angles.

Let ABC be any triangle ; any two of its angles together are less than two right angles.

Produce BC to D; and because Acd is the exterior angle of the triangle A B C, ACD is B greater (1. 16.) than the interior and opposite angle ABC; to each of these add the angle A CB; therefore the angles A C D, A C B are greater than the angles ABC, ACB; but A CD, ACB are together equal (1. 13.) to two right angles; therefore the angles A B C, BCA are less than two right angles. In like manner, it may be demonstrated, that B AC, AC B, as also C A B, A B C, are less than two right, angles. Therefore any two angles, &c. Q. E. D.

PROP. XVIII. THEOR. The greater side of every triangle is opposite to the greater angle.

Let A B C be a triangle, of which the side A č is greater than the side A B; the angle A B C is also greater than the angle B C A.

Because A c is greater than a B, make (1. 3.) AD equal to A B, and join BD; and because ADB is the exterior angle of the triangle B DC, it is greater (1. 16.) than the interior and opposite angle DCB; but A D B is equal (1. 5.) to ABD, because the side A B is equal to the side AD; therefore the angle ABD is likewise

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