The two sides FH, HC are equal to the two sides GH, HC, each to each; And the base CF is equal to the base CG (Def. 15); Therefore the angle CHF is equal to the angle CHG (I. 8), .. adjacent and they are adjacent angles. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle, and the straight line which stands on the other is called a perpendicular to it (Def. 10). Therefore, from the given point C, a perpendicular has been drawn to the given straight line AB. Q. E. F. Proposition 13.-Theorem. The angles which one straight line makes with another upon 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 angles shall either be two right angles, or shall together be equal to two right angles. PROOF. If the angle CBA be equal to the angle ABD, each of them is a right angle (Def. 10). angles CHF, CHG are equal. Make But if the angle CBA be not equal to the angle ABD, from the point B, draw BE at right angles to CD (I. 11). Therefore the angles CBE, EBD, are two right angles. Now the angle CBE is equal to the two angles CBA, ABE; 2 EBD = to each of these equals add the angle EBD. 4 CBE = a right . Therefore the angles CBE, EBD, are equal to the three angles CBA, ABE, EBD (Ax. 2). /CBE+ 2 EBD = < CBA + < ABE + Again, the angle DBA is equal to the two angles DBE, ZEBD, alEBA; to each of these equals add the angle ABC. Therefore the angles DBA, ABC, are equal to the three angles DBE, EBA, ABC (Ax. 2). SODBA + ABC DBE + EBA + 4 ABC. . ABE But the angles CBE, EBD have been shown to be equal to the same three angles; And things which are equal to the same thing are equal to one another; Therefore the angles CBE, EBD, are equal to the angles DBA, ABC (Ax. 1). But the angles CBE, EBD are two right angles. Therefore the angles DBA, ABC, are together equal to two right angles (Ax. 1). Therefore, the angles which one straight line, &c. Q. E. D. Proposition 14.-Theorem. 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 A E B D straight lines BC, BD, upon the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles. BD shall be in the same straight line with BC. For if BD be not in the same straight line with BC, let BE be in the same straight line with it. PROOF.-Because CBE is a straight line, and AB meets it in B. Therefore the adjacent angles ABC, ABE are together equal to two right angles (I. 13). But the angles ABC, ABD, are also together equal to two right angles (Hyp.); Therefore the angles ABC, ABE, are equal to the angles ABC, ABD (Ax. 1). Take away the common angle ABC. The remaining angle ABE is equal to the remaining angle = 2 ABD. ABD (Ax. 3), the less to the greater, which is impossible; Therefore BE 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. Therefore BD is in the same straight line with BC. Proposition 15.-Theorem." If two straight lines cut one another, the vertical, or opposite angles shall be equal. Let the two straight lines AB, CD cut one another in the point E. The angle AEC shall be equal to angle DEB, and the angle CEB to the angle AED. A E B < CEA + 4 AED == 2 right PROOF. Because the straight line AE makes with CD, the angles CEA, AED, these angles are together equal to two right angles angles. (I. 13). < AED + Again, because the straight line DE makes with AB the angles AED, DEB, these also are together equal to two right DEB = angles (I. 13). But the angles CEA, AED have been shown to be together equal to two right angles, Therefore the angles CEA, AED are equal to the angles AED, DEB (Ax. 1). Take away the common angle AED. 2 right angles. The remaining angle CEA is equal to the remaining angle CEA DEB (Ax. 3). In the same manner it can be shown that the angles CEB, AED are equal. Therefore, if two straight lines, &c. Q. E. D. COROLLARY 1.-From this it is manifest that if two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles. COROLLARY 2.-And, consequently, that all the angles made by any number of lines meeting in one point are together equal to four right angles, provided that no one of the angles be included in any other angle. = 4 DEB. Make AE EC. and EF BE. <BAE Proposition 16.—Theorem. If one side of a triangle be produced, the exterior angle shall be 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 ACD shall be greater than either of the interior opposite angles CBA, BAC. A P D CONSTRUCTION.-Bisect AC in E (T. 10). Join BE, and produce it to F, making EF equal to BE (I. 3), and join FC PROOF. Because AE is equal to EC, and BE equal to EF (Const.), AE, EB are equal to CE, EF, each to each ; And the angle AEB is equal to the angle CEF, because they are opposite vertical angles (I. 15). Therefore the base AB is equal to the base CF (I. 4); And the remaining angles to the remaining angles, each to each, to which the equal sides are opposite. Therefore the angle BAE is equal to the angle ECF ECF. (I. 4). .. ¿ACD BAE. But the angle ECD is greater than the angle ECF (Ax. 9); Therefore the angle ACD is greater than the angle BAE. In the same manner, if BC be bisected, and the side AC be produced to G, it may be proved that the angle BCG (or its equal ACD), is greater than the angle ABC. Therefore, if one side, &c. Q. E. D. Proposition 17.-Theorem. Any two angles of a triangle are together less than two right angles. Let ABC be any triangle. Any two of its angles together shall be less than two right angles. CONSTRUCTION.--Produce BC to D. PROOF. Because ACD is the exterior angle of the triangle ABC, it is greater than the interior and opposite angle ABC (I. 16). To each of these add the angle АСВ. B < ACD > 4 ABC. Add toeach 2. ACB. Therefore the angles ACD, ACB are greater than the ABC + angles ABC, ACB (Ax. 4). < ACB < 2 right But the angles ACD, ACB are together equal to two angles. right angles (I. 13); Therefore the angles ABC, ACB are together less than two right angles. In like manner, it may be proved that the angles BAC, ACB, as also the angles CAB, ABC are together less than two right angles. Therefore, any two angles, &c. Q. E. D. Proposition 18.-Theorem. The greater side of every triangle is opposite the greater angle. Let ABC be a triangle, of which the side AC is greater AC > AB. than the side AB. The angle ABC shall be greater than the angle BCA. CONSTRUCTION.-Because AC is greater than AB, make AD equal to AB (I. 3), and join BD. B PROOF.-Because ADB is the exterior angle of the triangle BDC, it is greater than the interior and opposite angle BCD (I. 16). But the angle ADB is equal to the angle ABD; the triangle BAD being isosceles (I. 5), Therefore the angle ABD is greater than the angle BCD (or ACB). Much more then is the angle ABC greater than the angle АСВ. Therefore, the greater side, &c. Q. E. D. Make |