F * For, take the equal Right Lines EA, EB, CE, DE; and, through E any how draw the Right Line GEH, and join AD, CB; and from the Point'F, let there be drawn Fa, F G, F D, FC, F H, FB: Then, because two Right Lines A E, ED, are equal to two Right Lines CE, E B, and they contain * the * 15.30 equal Angles A ED, CEB; the Base A D shall be ++ 4. ti equal to the Bale C B, and the Triangle A E D equal to the Triangle CEB; and so, likewise, is the Angle DAE equal to the Angle EBC; but the Angle A E G is * equal to the Angle B EH; therefore AGE, B E H, 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 so the other Sides of the one will be f equal to the other Şides of [ 26. t. the other. Therefore G E is equal to E H, and AG to BH; and since A E is equal to E B, and F E is common and at Right Angles, the Base A F shall be 7.4. 16 equal to the Base FB: For the same Reason, likewise, thall C F be equal to FD. Again, because A D is equal to C B, and AF to F B, the two Sides FA, AD, will be equal to the two Sides F B, BC, each to each ; but the Base D F has been proved equal to to the Angle FBC: Moreover, AG has been proved equal to BH; but F B also, is equal to AF, there. fore the two Sides FA, AG, are equal to the two Sides F B, BH; and the Angle F A G is equal to the Angle FBH, as has been demonstrated; wherefore tbe Base G F is f equal to the Base FH. Again, because G E has been proved equal to EH, and EF is cominon, the two Sides G E, EF, are equal to the two Sides A E, EF; but the Base H F is equal to the Base FG; therefore the Angle G E F is * equal to the Angle HEF; and so both the Angles G EF, HEF, are Right Angles : Therefore, FE makes Right Angles with G H, which is any how drawn thro' E, After the same manner, we demonstrate, that E F 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 Anglestbis. to all Right Lines drawn to it in the Plane. Therefore F E is at Right Angles to a Plane drawn thro’ the 02 Right 1. 1 Right Line A B, CD. Wherefore, if to two Right Lines cutting one another, a third stands at Right Angles in the common Section, it shall be also at Right Angles to the Piane drawn through the said Lines; which was to be deinonstrated. PROPOSITION V. THEOREM. if to three Right Lines touching one another, a fourth stands at Right Angles in their common Section, those three Rigbt Lines shall be in one and the same Plane. L E T the Right Line A B stand at Right Angles, in the Point of Contact B, to the three Right Lines B C, B D, B E. I say, B C, B D, B E, are in one and the same Plane. For, if they are not, let D B, BE, be in one Plane, and B C above it; and let the Plane passing thro’A B, * 3 of ibis. B C, be produced, and it will * make the common Section, with the other Plane, a strait 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 since A B stands at Right Angles to B D and BE, it shall +4 of ebis, be † at Right Angles to a Plane drawn through BE, Defi's. DB; and so A B shall inake Right Angles with all Right Lines touehing it that are in the fame Plane: But B F, being in the said Plane, touches it; wherefore the Angle A B F is a Right Angle: But the Angle A B C (by the Hyp.) is also a Right Angle; therefore the Angle A BF is equal to the Angle A B C, and they are boih in the same Plane, which cannot be; and so the Right Line B C, is not above the Plane paffing through EB and B D. Wherefore the three Right Lines BC, BD, BE, are in one and the same Plane. Therefore, if to three Right Lines, touching one another, a fourth stands at Right Angles in their common Section, ihose three Right Lines fall be in one and the fame Plane ; which was to be demonstrated. PRO PROPOSITION VI. THEORE M. if two Right Lines be. perpendicular to one and the fame Plane, those Right Lines are parallel to one anoiber. LET two Right Lines AB, CD, be perpendicular to one and the same Plane. I say, A B is parallel to CD. For, let them meet the Plane in the Points B, D ; and join the Right Line B D, to which let D E be drawn in the fame Plane, at Right Angles, make D E equal to A B ; and join B E, A E, A D. Then because AB is at Right Angles to the aforesaid Plane, it shall be * at Right Angles to all Right * Def. 3. of Lines, touching it, drawn in the Plane ; but A B biso touches BD, BE, which are in the said Plane ; therefore each of the Angles A BD, ABE, is a Right Angle. So, for the fame Reason, likewise, is each of the Angles C DB, C D E, a Right Angle. Then, because A B is equal to D Е, and B D is common; the iwo Sides A B, BD, shall be equal to the two Sides ED, DB; but they contain Right Angles : Therefore the Base A D ist equal to the Bale B E.+ 4. 1. Again, because A B is equal to D Е, and A D to BE; the two Sides A B, B E, are equal to the two Sides ED, DA; but A E, their Bale, is common; wherefore the Angle A B Eis I is equal to the Angle EDA But A B E is a Right Angle ; therefore E D A is also a Right Angle ; and fo E D is perpendicular to DA: But it is alio perpendicular to B D and DC; therefore E D is at Right Angles, in the Point of Contact, to three Right Lines B D, DĄ, DC: Wherefore these three last Right Lines are * in one Plane. But * 5 of Ibis. BD, D A, are in the fame Plane as A B is : for every Triangle is t in the fame Plane; therefore it is necel + 2 of ibis. fary, that AB, BD, DC, be in one Plane. But both the Angles A BD, BD C, are Right Angles ; wherefore AB is I parallel to CD. Therefore, if 1 28. 1. two Right Lines be perpendicular to one and the same Plane, those Right Lines are parallel to one another ; which was to be demonstrated. PROPOSITION VIT. THEOREM. taken in both of them, the Right Line joining ET AB, CD, be two parallel Right Lines, in which are taken any Points, E, F. I say, 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 same, if possible, as E GF, through which let fome. Plane be drawn, whose Section, with the Plane in which the * 3 of ibis. Parallels are, let * be the Right Line E F; then the two Right Lines E G F, EF, will include a Space, | Ax,10.3 which is + absurd : Therefore, a Right Line, drawn from the Point E to the Point F, is not elevated above the Plane; and, consequently, it must be in that passing 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 those Points ball be in the fame Plane as the Parallels are ; which was to be demonstrated. PROPOSITION VIII. THEOREM. is perpendicular to fome Plane ; i ben shall tbe other be perpendicular to the fame Plane. See sbe Fig. of Prup. VI. LET A B, CD, be two paraltel Right Lines, one of which, as AB, is perpendicular to fome Plane, I say, the other, CD, is also perpendicular to the same Plane. For, let AB, CD, meet the Plane in the Points B, D; and let B D be joined ; then A B, C D, B D, are " of Ibis. in one Plane. Let D E be drawn in the other Plane, Plane, at Right Angles to BD, and make DE equal † 29. I. |