1. The base AD is equal to the base BC, and the angle DAE to the angle EBC (I. 4). B But the angle AEG is equal to the angle BEH (I. 15). Therefore the two triangles AEG and BEH have two angles of the one equal to two angles of the other, each to each, and the sides AE and EB, adjacent to the equal angles, equal to one another, therefore 2. Their other sides are equal, that is, GE is equal to EH, and AG to BH (I. 26); and because AE is equal to EB, FE common, and the angle AEF equal to the angle BEF (hyp.), therefore 3. The base AF is equal to the base FB (Ț. 4). For the same reason, CF is equal to FD. to BC, and AF to FB, the two sides FA two FB and BC, each to each. But the equal to the base FC, therefore And because AD is equal and AD are equal to the base DF has been proved · 4. The angle FAD is equal to the angle FBC (I. 8). But it has been proved that GA is equal to BH, and also AF to FB; therefore 5. FA and AG are equal to FB and BH, each to each; and the angle FAG has been proved equal to the angle FBH, therefore 6. The base GF is equal to the base FH (I. 4). But it has been proved that GE is equal to EH, and EF is common. Therefore GE and EF are equal to HE and EF, each to each; and the base GF is equal to the base FH, therefore 7. The angle GEF is equal to the angle HEF (I. 8), and each of these angles is a right angle (I. Def. 10); therefore 8. The straight line FE makes right angles with GH, that is, with any straight line drawn through E in the plane of AB and CD. In like manner, it may be proved that FE makes right angles with every other straight line meeting it in that plane. But a straight line is at right angles to a plane when it makes right angles with every straight line meeting it in that plane (XI. Def. 3); therefore 9. EF is at right angles to the plane of AB and CD. Wherefore, if a straight line, &c. Q.E.D. PROPOSITION 5.-Theorem. If three straight lines meet all in one point, and a straight line stands at right angles to each of them at that point; these three straight lines are in the same plane. Let the straight line AB stand at right angles to each of the straight lines BC, BD, and BE at the point B where they meet. Then the straight lines BC, BD, and BE are in the same plane. E Construction. If they be not in the same plane, let, if possible, BD and BE be in one plane, and BC out of it. Let a plane pass through AB and BC, and let the straight line BF be the common section of this plane, and the plane of BD and BE (XI. 3). Demonstration. Because the three straight lines AB, BC, and BF are all in one plane (hyp.), viz., that which passes through AB and BC: and AB stands at right angles to each of the straight lines BD and BE; therefore AB is at right angles to the plane in which they are (XI. 4); wherefore 1. AB makes right angles with every straight line meeting it in that plane (XI. Def. 3); but BF, which is in that plane, meets it; therefore 2. The angle ABF is a right angle. But the angle ABC is also a right angle (hyp.); therefore 3. The angle ABF is equal to the angle ABC, and they are both in the same plane; which is impossible (1 Ax. 9); therefore Therefore 4. The straight line BC is not out of the plane of BD and BE; The three straight lines BC, BD, and BE are in the same plane. Therefore, if three straight lines, &c. Q.E.D. PROPOSITION 6.-Theorem. If two straight lines be at right angles to the same plane, they are parallel. Let the straight lines AB and CD be at right angles to the same plane. Then AB is parallel to CD. therefore Construction. Let the straight lines meet the plane in the points B and D. Join BD, and draw DE at right angles to it in the same plane (I. 11). Make DE equal to AB (I. 3), and join BE, AE, and AD. Demonstration. Because AB is perpendicular to the plane; 1. AB makes right angles with every straight line meeting it in that plane (XI. Def. 3) ; but BD and BE in that plane each meet AB; therefore 2. Each of the angles ABD and ABE is a right angle; for the same reason 3. Each of the angles CDB and CDE is a right angle. Because AB is equal to DE, and BD common, the two sides AB and BD are equal to the two ED and DB, each to each, and they contain right angles; therefore 4. The base AD is equal to the base BE (I. 4). • Again, because AB is equal to DE, and BE to AD; AB and BE are equal to ED and DA, each to each; and, in the triangles ABE and EDA, the base AE is common; therefore 5. The angle ABE is equal to the angle EDA (I. 8), But ABE is a right angle, therefore 6. EDA is also a right angle, and ED perpendicular to ᎠᎪ ; but it is also perpendicular to each of the two BD and DC. Therefore ED is at right angles to each of the three straight lines BD, DA, and DC at the point Ď where they meet; wherefore 7. These three straight lines are all in the same plane (XI. 5). But AB is in the plane of BD and DA, because any three straight lines which meet in three points are in one plane (XI. 2); therefore 8. AB, BD, and DC are in one plane, and each of the angles ABD and BDC is a right angle; therefore 9. AB is parallel to CD (I. 28). Wherefore, if two straight lines, &c. Q.E.D. PROPOSITION 7.-Theorem. If two straight lines be parallel, the straight line drawn from any point in the one to any point in the other, is in the same plane with the parallels. Let AB and CD be parallel straight lines, and E any point in the one, and F any point in the other. Then the straight line which joins E and F is in the same plane with the parallels. E D Construction. If this straight line be not in the same plane with them, let it, if possible, be out of the plane, as EGF. In the plane AD of the parallels, draw the straight line EHF from E to F. Demonstration. Because the two straight lines EHF and EGF have the same extremities and do not coincide, E с Wherefore H 1. EHF and EGF include a space between them, which is impossible (I. Ax. 10); therefore D 2. The straight line joining the points E and F is not out of the plane of the parallels AB and CD. 3. It is in that plane. Therefore, if two straight lines, &c. Q.E.D. PROPOSITION 8.-Theorem. If two straight lines be parallel, and one of them be at right angles to a plane, the other is also at right angles to the same plane. Let AB and CD be two parallel straight lines, and let one of them AB be at right angles to a plane. Then the other CD is also at right angles to the plane. Construction. Let AB and CD meet the plane in the points B and D, and join BD; therefore 1. The straight lines AB, CD, and BD are in one plane (XL. 7). In the plane BDE draw DE at right angles to BD (I. 11); make DE equal to AB (I. 3), and join BE, AE, and AD. |