straight lines AE, ED are equal to angle FBC: Again it was proved the two BE, EC, and that they con- that AG is equal to BH, and also AF tain equal angles (10. Ax. 1.) AED, to FB; FA, then, and AG, are equal BEC, the base AD is equal (4. 1.) to to FB and BH, and the angle FAG the base BC, and the angle DAE to has been proved equal to the angle the angle EBC: And the angle AEG FBH; therefore the base GF is equal is equal to the angle BEH(10. Ax. 1.); (4. 1.) to the base FH: Again, betherefore the triangles AEG, BEH cause it was proved that GE is equal have two angles of one equal to two to EH, and EF is common; GE, EF angles of the other, each to each, and are equal to HE, EE; and the base the sides AF, EB, adjacent to the GF is equal to the base FH: thereequal angles, equal to one another; fore the angle GEF is equal (8. 1.) to wherefore they shall have their other the angle HEF; and consequently sides equal (26. 1.): GE is therefore each of these angles is a right (10. equal to EH, and AG to BH: And def.) angle. Therefore FE makes because AE is equal to EB, and FE right angles with GH, that is, with common and at right angles to them, any straight line drawn through E in the base AF is equal (4. 1.) to the base the plane passing through AB; CD. FB; for the same reason, CF is equal In like manuer, it may be proved, to FD: and because that FE makes right angles with every AD is equal to BC, straight line which meets it in that and AF to, FB, the plane. But a straight line is at right two sides FA, AD, AK с angles to a plane when it makes right are equal to the two G angles with every straight line which FB, BC, each to each; meets it in that plane (3. def. 11.): and the base DF was H Therefore EF is at right angles to the proved equal to the D plane in which are AB, CD. Wherebase FC; therefore fore, if a straight line, &c. Q. E. D. the angle FAD is equal (8. 1.) to the E F PROP. V. THEOREM. If three straight lines meet all in one point, and a straight line stands at right angles to each of them in that point; these three straight lines are in one and the same plane. Let the straight line AB stand at right angles to each of the straight right angles to each of the straight lines BD, BE, it is also at right angles lines BC, BD, BE, in B the point (4. 11.) to the plane passing through where they meet ; BC, BD, BE are them; and therefore makes right anin one and the same plane. gles (3. def. 11.) with every straight If not, let, if it be possible, BD and line meeting it in BE be in one plane, and BC be above that plane ; but BF it; and let a plane pass through AB, which is in that BC, the common section of which plane meets it : with the plane, in which BD and BE Therefore the angle are, shall be a straight (3. 11.) line; ABF is a right anlet this be BF: Therefore the three gle; but the angle straight lines AB, BC, BF, are all in ABC, by the hypoone plane, viz. that which passes thro' thesis, is also a B AB, BC; and because AB stands at right angle; there D fore the angle ABF is equal to the BE: Wherefore the three straight angle ABC, and they are both in the lines BC, BD, BE are in one and the same plane, which is impossible : same plane. Therefore, if three Therefore the straight line BC is not straight lines, &c. Q. E. D. above the plane in which are BD and PROP. VI. THEOREM. If two straight lines be at right angles to the same plane, they shall be parallel to one another. Let the straight lines AB, CD be base AD is equal (4. 1.) to the base at right angles to the same plane; BE: Again, because AB is equal to AB is parallel to CD. DE, and BE to AD; AB, BE are Let them meet the plane in the equal to ED, DA; and in the tripoints B, D, and draw the straight angles ABE, EDA, the base AE is line BD, to which draw DE at right common: therefore the angle ABE is angles, in the same plane; and make equal (8. 1.) to the angle EDA.: But DE equal to AB, and join BE, AE, ABE is a right angle; therefore EDA AD. Then, because AB A с is also a right angle, and ED perpenis perpendicular to the dicular to DA: But it is also perpenplane, it shall make dicular to each of the two BD, DC: right (3.def. 11.) angles Wherefore ED is at right angles to with every straight line each of the three straight lines BD, which meets it, and is B DA, DC in the point in which they in that plane: But BD, meet: Therefore these three straight BE, which are in that E lines are all in the same plane (5. 11.): plane, do each of them But AB is in the plane in which are meet AB. Therefore each of the an- BD, DA, because any three straight gles ABD, ABE is a right angle: For lines which meet one another are in the same reason, each of the angles one plane (2. 11.): Therefore AB, BD, CDB, CDE is a right angle: And be- DC are in one plane: And each of the cause AB is equal to ĎE, and BD angles ABD, BDC is a right angle ; common, the two sides AB, BD are therefore AB is parallel (28. 1.) to equal to the two ED, DB ; and they CD. Wherefore, if two straight lines, contain right angles; therefore the &c. Q. E. D., PROP VII. 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, CD be parallel straight draw the straight line EHF from lines, and take any point E in the one, E to F; and and the point F in the other : The since EGF also A I B straight line which joins E and F is is a straight in the same plane with the parallels. line, the two H н If not, let it be, if possible, above straight lincs the plane, as EGF; and in the plane EHF, EGF inABCD in which the parallels are, clude a space D 個 between them, which is impossible : CD are, and is therefore in that plane. (10.Ax. 1.) Therefore the straight line Wherefore, if two straight lines, &c. joining the points E, F is not above Q. E. D. the plane in which the parallels AB, PROP. VIII. THEOREM. If two straight lines be parallel, and one of them is at right angles to a plane ; the other also shall be at right angles to the same plane. Let AB, CD be two parallel straight therefore the base AD is equal (4. lines, and let one of them AB be at 1.) to the base BE: Again, because right angles to a plane; the other CD AB is equal to DE, and BE to AD; is at right angles to the same plane. the two AB, BE are equal to the two Let AB, CD meet the plane in the ED, DA; and the base AE is compoints B, D, and join BD: Therefore mon to the triangles ABE, EDA; (7.11.) AB, CD, BD are in one plane. wherefore the angle ABE is equal In the plane to which AB is at right (8. 1.) to the angle EDA: And ABE angles, draw DE at right angles to is a right angle; and therefore EDA BĎ, and make DE equal to AB, and is a right angle, and ED perpendicular join BE, AE, AD. And because AB to DĂ: But it is also perpendicular is perpendicular to the plane, it is per- to BD; therefore ED is perpendicular pendicular to every straight line which (4. 11.) to the plane which passes meets it, and is in that plane:(3. Def, through BD, DA, and shall (3. def. 11.) Therefore each of the angles 11.) make right angles with every ABD, ABE, is a right angle: And straight line meeting it in that plane: because the straight line BD meets But DC is in the plane passing through the parallel straight lines AB, CD, the BD, DA, because all three are in the angles ABD, CDB are together equal plane in which are the parallels AB, (29. 1.) to two right angles: And ABD CD: Wherefore ED is at right angles is a right angle; therefore also CDB to DC; and therefore CD is at right is a right angle, and CD angles to DE: But CD is also at perpendicular to BD: right angles to DB; CD then is at And because AB is equal A right angles to the two straight lines to DE, and BDcommon, DE, DB in the point of their interthe two AB, BD are section D; and therefore is at right equal to the two ED, angles (4. 11.) to the plane passing DB, and the angle ABDB through DE, DB, which is the same is equal to the angle plane to which AB is at right angles. EDB, because each of Therefore, if two straight lines, &c. them is a right angle ; Q. E. D. Two straight lines which are each of them parallel to the same straight line, and not in the same plane with it, are parallel to one another. Let AB, CD be each of them par- draw in the plane passing through EF, allel to EF, and not in the same plane AB, the straight line GH at right with it; AB shall be parallel to CD. angles to EF ; and in the plane passe In EF take any point G, from which ing through EF, CD, draw GK at right angles to the same EF. And to the plane HGK. For the same because EF is perpendicular both to reason, CD is likewise at right angles GH and GK,EF H to the plane HGK. Therefore AB, is perpendicular A B CD are each of them at right angles (4. 11.) to the to the plane HGK. But if two straight plane HGK pass lines are at right angles to the same I I ing thro' them : plane, they shall be parallel (6. 11.) And EF is par-c D to one another. Therefore AB is parallel to AB: allel to CD. Wherefore, two straight therefore AD is at right angles (8.11.) lines, &c. Q. E. D. K PROP. X. THEOREM. one If two straight lines meeting one another be parallel to two others that meet one another, and are not in the same plane with the first two ; the first two and the other two shall contain equal angles. Let the two straight lines AB, BC, and not in the same plane with it, are which meet one another be parallel parallel (9. 11.) to the two straight lines DE, ÈF that io another. B meet one another, and are not in the Therefore AD is same plane with AB, BC. The angle parallel to CF; and A ABC is equal to the angle DEF. it is equal (1. Ax. Take BA, BC, ED, EF, all equal 1.) to it, and AC, to one another : and join AD, CF, BE, DF join them, to- E AC, DF; because BA is equal and wards the same parallel to ED, therefore AD is (33. parts: and therefore D F 1.) both equal and parallel to BE. (33. 1.) AC is equal and parallel to For the same reason, CF is equal and DF. And because AB, BC are equal parallel to BE. Therefore AD and to DE, EF, and the base AC to the CF are each of them equal and par- base DF; the angle ABC is equal allel to BE. But straight lines that (8. 1.) to the angle DEF. Therefore, are parallel to the same straight line, if two straight lines, &c. Q. E. D. PROP. XI. PROBLEM. To draw a straight line perpendicular to a plane, from a given point above it. Let A be the given point above the BC; and from the point A draw AF plane BH; it is required to draw perpendicular to DE; and through from the point A a straight line pere F draw (31., 1.) GH E A pendicular to the plane BH. parallel to BC: And In the plane draw any straight line because BC is at y el H BC, and from the point A draw AD right angles to ED perpendicular to BC. If then AD be and DA, BC is at also perpendicular to the plane BH, right angles (4. 11.) the thing required is already done; to the plane pass D but if it be not, from the point D ing through ED, DA. And GH is draw, (11. 1.) in the plane BH, the parallel to BC; but, if two straight straight line DE, at right angles to lines be parallel, one of which is at U B right angles to a plane, the other shall a straight line stands at right angles be at right (8. 11.) angles to the same to each of two straight lines in the plane: wherefore GH is at right angles point of their intersection, it shall also to the plane through ED, DA, and is be at right angles to the plane passing perpendicular (3. def. 11.) to every through them. But the plane passe straight line meeting it in that plane. ing through ED, GH, is the plane But ÅF, which is in the plane through BÉ; therefore AF is perpendicular ED, DA meets it: Therefore GH is to the plane BH: therefore from perpendicular to AF; and conse- the given point A, above the plane quently AF is perpendicular to GH; BH, the straight line AF is drawn perand AF is perpendicular to DE: pendicular to that plane : Which was Therefore AF is perpendicular to each to be done. of the straight lines GH, DE. But if PROP. XII. PROBLEM. To erect a straight line at right angles to a given plane, from a point in the plane. Let A be the point given in the AD, CB are two parallel straight lines, plane; it is required to erect a straight and one of them BC is at right angles line from the point A at DB to the given plane, the other AĎ is right angles to the plane. also at right angles to it. (8. 1.) From any point B above Therefore a straight line has been the plane draw (11. 11.) erected at right angles to a given plane BC perpendicular toit; and from a point given in it. Which was А с from A draw (31. 1.) AD to be done. parallel to BC. Because, therefore, PROP. XIII. THEOREM. From the same point in a given plane, there cannot be two straight lines at right angles to the plane, upon the same side of it. And there can be but one perpendicular to a plane from a point above the plane: For, if it be possible, let the two given plane, it shall make right angles straight lines AB, AC, be at right with every straight line meeting it in angles to a given plane from the same that plane. But DAE, which is in point A in the B that plane, meets CA; therefore CAE planc, and upon is a right angle. For the same reason the same side of BAE is a right angle. Wherefore it; and let a plane the angle CAE is equal to the angle pass through BA, BAE; and they are in one plane, AC; the common which is impossible. Also, from a section of this with the given plane is point above a plane, there can be but a straight (3. 11.) line passing through one perpendicular to that plane ; for, A: Let DAE be their common sec- if there could be two, they would be tion: Therefore the straight lines, parallel (6. 11.) to one another, which AB, AC, DAE are in one plane : And is absurd. Therefore, from the same because CA is at right angles to the point, &c. Q. E. D. D |