For, take the equal Right Lines E A, E B, CE, DE; and through E any how draw the Right Line GEH, and join A D, CB; and from the Point F, let there be drawn FA, F G, FD, FC, F H, FB: Then, because two Right Lines A E, E D, are equal * 8. . I. to two Right Lines CE, EB, and they contain *the* 15. 16 equal Angles A ED, CEB; the Bafe A D fhall be †† 4. 1. equal to the Bafe C B, and the Triangle A E D equal to the Triangle CEB; and fo, likewife, is the Angle DAE equal to the Angle E BC; but the Angle AEG is equal to the Angle BEH; therefore AGE, BEH, are two Triangles, having two Angles of the one equal to two Angles of the other, each to each, and one Side A E equal to one Side E B; viz. those that are at the equal Angles; and fo the other Sides of the one will be equal to the other Sides of 26. 1. the other. Therefore G E is equal to E H, and AG to BH; and fince AE is equal to E B, and F E is common and at Right Angles, the Bafe A F fhall be + equal to the Bafe F B: For the fame Reason, likewife, fhall C F be equal to F D. Again, because A D is equal to C B, and AF to F B, the two Sides F A, AD, will be equal to the two Sides F B, BC, each to each; but the Bafe DF has been proved equal to the Bafe FC. Therefore the Angle FA D is equal to the Angle F BC: Moreover, AG has been proved equal to В H; but F B alfo, is equal to A F, therefore the two Sides F A, AG, are equal to the two Sides F B, BH; and the Angle FAG is equal to the Angle FBH, as has been demonftrated; wherefore the Bafe GF is + equal to the Bafe FH. Again, because GE has been proved equal to E H, and E F is common, the two Sides G E, E F, are equal to the two Sides HE, EF; but the Bafe H F is equal to the Base F G; therefore the Angle GEF is equal to the Angle HEF; and fo both the Angles G EF, HEF, are Right Angles: Therefore, F E makes Right Angles with GH, which is any how drawn thro' E. After the fame manner we demonftrate, that EF is at Right Angles to all Right Lines that are drawn in a Plane to it; but a Right Line is at Def. 3. of Right Angles to a Plane, when it is at Right Angles this. to all Right Lines drawn to it in the Plane. Therefore F E is at Right Angles to a Plane drawn thro' the Q 2 Right Right Line A B, CD. Wherefore, if to two Right Lines cutting one another, a third fands at Right Angles in the common Section, it shall be alfo at Right Angles to the Piane drawn through the faid Lines; which was to be demonftrated. PROPOSITION V. THEOREM. If to three Right Lines touching one another, a fourth ftands at Right Angles in their common Section, those three Right Lines fhall be in one and the fame Plane.. ET the Right Line A B ftand at Right Angles, to the LET either For, if they are not, let D B, B E, be in one Plane, and BC above it; and let the Plane paffing thro' A B, * of this. BC, be produced, and it will make the common Section, with the other Plane, a ftrait Line, which let be B F; then three Right Lines A B, BC, B F, are in one Plane drawn through A B, BC: And fince A B ftands at Right Angles to BD and BE, it shall +4 of this, bet at Right Angles to a Plane drawn through BE, Def. 3. DB; and fo A B fhall make † Right Angles with all Right Lines touehing it that are in the fame Plane: But B F, being in the faid Plane, touches it; wherefore the Angle ABF is a Right Angle: But the Angle ABC (by the Hyp.) is alfo a Right Angle; therefore the Angle A B F is equal to the Angle A B C, and they are both in the fame Plane, which cannot be; and fo the Right Line B C, is not above the Plane paffing through EB and B D. Wherefore the three Right Lines BC, BD, B E, are in one and the fame Plane. Therefore, if to three Right Lines, touching one another, a fourth ftands at Right Angles in their common Section, thofe three Right Lines fhall be in one and the fame Plane; which was to be demonftrated. PRO VI. PROPOSITION If two Right Lines be perpendicular to one and the LET two Right Lines A B, CD, be perpendicular to one and the fame Plane. I fay, AB is parallel to C D. For, let them meet the Plane in the Points B, D ; and join the Right Line B D, to which let DE be drawn in the fame Plane, at Right Angles, make DE equal to A B; and join BE, AE, AD. Then becaufe AB is at Right Angles to the aforefaid Plane, it fhall be at Right Angles to all Right * Def. 3. of Lines, touching it, drawn in the Plane; but A B bis. touches BD, BE, which are in the faid Plane; therefore each of the Angles A BD, ABE, is a Right Angle. So, for the fame Reafon, likewife, is each of the Angles CD B, C D E, a Right Angle. Then, becaufe A B is equal to D E, and B D is common; the two Sides A B, BD, fhall be equal to the two Sides ED, DB; but they contain Right Angles : Therefore the Bafe A D is equal to the Bafe BE. † 4. 1. Again, because A B is equal to D E, and AD to BE; the two Sides A B, BE, are equal to the two Sides ED, DA; but A E, their Bafe, is common ; wherefore the Angle A B Eis is equal to the Angle ED A.18. 1. But A B E is a Right Angle; therefore E D A is alfo a Right Angle; and fo ED is perpendicular to DA: But it is also perpendicular to B D and DC; therefore ED is at Right Angles, in the Point of Contact, to three Right Lines BD, DA, DC: Wherefore these three laft Right Lines are * in one Plane. But * BD, D A, are in the fame Plane as A B is: for every Triangle is in the fame Plane; therefore it is necef-+ 2 of this. fary, that AB, B D, DC, be in one Plane. But both the Angles A B D, BD C, are Right Angles; wherefore A B is parallel to CD. Therefore, ift 28. 1. two Right Lines be perpendicular to one and the fame Plane, thofe Right Lines are parallel to one another; which was to be demonstrated. 5 of this. +Ax, 10.1. PROPOSITION VII. THEOREM. If there be two parallel Lines, and any Points be taken in both of them, the Right Line joining thofe Points fhall be in the fame Plane as the Parallels are. LET AB, CD, be two parallel Right Lines, in which are taken any Points, E, F. I fay, a Right Line joining the Points E, F, is in the fame Plane as the Parallels are. For, if it be not, let it be elevated above the fame, if poffible, as EGF, through which let fome. Plane be drawn, whofe Section, with the Plane in which the 3 of this. Parallels are, let be the Right Line E F; then the two Right Lines E GF, EF, will include a Space, which is abfurd: Therefore, a Right Line, drawn from the Point E to the Point F, is not elevated above the Plane; and, confequently, it must be in that paffing through the Parallels A B, CD. Wherefore, if there be two parallel Lines, and any Points be taken in both of them, the Right Line joining thofe Points fall be in the fame Plane as the Parallels are; which was to be demonftrated. See she Fig. PROPOSITION VIII. THEOREM. If there be two parallel Right Lines, one of which is perpendicular to fome Plane; then fall the other be perpendicular to the fame Plane. of Prep. VI. LETA B, C D, be two parallel Right Lines, one of which, as A B, is perpendicular to fome Plane. I fay, the other, CD, is alfo perpendicular to the fame Plane. For, let A B, CD, meet the Plane in the Points B, D; and let BD be joined; then A B, C D, BD, are 7 of this. in one Plane. Let DE be drawn in the other Plane, Plane, at Right Angles to B D, and make DE equal to A B; and join BE, A E, AD: Then, fince A B is perpendicular to the Plane, it will be perpendicu- * Def. 3. lar to all Right Lines touching it, that are drawn in the fame Plane; therefore, each of the Angles A BD, A BE, is a Right Angle. And fince the Right Line BD falls on the Right Lines A B, CD; the Angles A BD, CD B, fhall be † equal to two Right Angles: † 29. &. 'Therefore, the Angle CDB is alfo a Right Angle; and fo CD is perpendicular to DB. And fince A B is equal to DE, and B D is common; the two Sides AB, BD, are equal to the two Sides ED, D B. But the Angle ABD is equal to the Angle E DB; for each of them is a Right Angle; therefore the Bafe AD is equal to the Bafe B E. Again, fince A B is 14. 1. equal to DE, and B E to AD; the two Sides A B, -BE, fhall be equal to the two Sides E D, D A, each to each: But the Bafe A E is common; wherefore, the Angle A BE is equal to the Angle EDA: But 8. 1. the Angle A B E is a Right Angle; therefore EDA is alfo a Right Angle, and ED is perpendicular to DA: But it is likewife perpendicular to D B; therefore ED fhall alfo be + perpendicular to the Plane + 4 of this. paffing through BD, DA, and, likewise, fhall be at Def. 3. Right Angles to all Right Lines, drawn in the faid Plane that touch it. But D C is in the Plane paffing through BD, DA, because A B, BD, are * in that * 2 of this Plane; and DC is + in the fame Plane that A B and † 7 of this. BD are in; wherefore ED is at Right Angles to DC, and fo CD is at Right Angles to D E, as alfo to D B. Therefore, CD ftands at Right Angles, in the common Section D, to two Right Lines DE, D B, mutually cutting one another; and, accordingly, is at Right Angles to the Plane paffing through D E, D B ; which was to be demonftrated. |