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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 centre 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 HG, and HC 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 CHF 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.

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The angles which one straight line makes with another upon the one side of it, are either two right angles, or 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 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

the point B draw BE at right angles (1. 11.) to CD; therefore the angles CBE, E B D, are two right angles (Def. 10.); D and because CBE is equal to

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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, EBD. Again, because the angle D BA is equal to the two angles DBE, EBA, add to these equals the angle ABC; therefore the angles D B A, ABC are equal to the three angles D B E, E BA, A B C ; but the angles CBE, EBD 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 CBE, E BD are equal to the angles DBA, ABC; but CBE, 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 AB, let the two straight 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.

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

therefore, because the straight line AB makes angles with the straight lines CBE, upon one side of it, the angles ABC, ABE are together equal (1. 13.) to two right angles; but the angles ABC, ABD are likewise together equal to two right angles (Hyp.); therefore the angles CBA, A BE are equal to the angles CBA, ABD. Take away the common angle ABC, the remaining angle ABE is equal (Ax. 3.) to the remaining angle A B D, the less to the greater, which is impossible; therefore BE 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.

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

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

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Because the straight line AE makes with CD the angles CE A, A E D, these angles are together equal (1. 13.) to two right angles. Again, because the straight line C DE makes with AB the angles A ED, DEB these also are together equal (1. 13.) to two right angles; and CEA, AED have been demonstrated to be equal to two right angles; wherefore the angles CEA, A E D are equal to the angles AED, DE B. Take away the common angle A E D, and the remaining angle CEA is equal (Ax. 3.) to the remaining angle DEB. In the same manner it can

be demonstrated that the angles CE B, AED 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.

PROP. XVI. THEOR.

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

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Let ABC be a triangle, and let its side BC be produced to D, the exterior angle A CD is greater than either of the interior opposite angles CBA, BA C. Bisect (1. 10.) AC in E, join BE and produce it to F, and make EF equal to BE; B join also FC, and produce

AC to G.

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Because A E is equal to EC and BE to EF; AE, E B are equal to CE, EF, 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 CFE, 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 ECF; but the angle E CD is greater than the angle ECF; therefore the angle ACD is greater than BA E. In the same manner, if the side BC be

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

angle ABC.

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Any two angles of a triangle are together less than two right angles.

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

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

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