LXM. For the fame Reafon, the Angle DEF is equal to the Angle MXN, and the Angle GHK to the Angle NXL. Therefore the three Angles ABC, DEF, GHK, are equal to the three Angles LXM, MXN, NXL. But the three Angles LX M, MXN, NX L, are * equal to four Right Angles :* Cor. 15. * And fo the three Angles ABC, DEF, GHK, fhall alfo be equal to four Right Angles; but they are put lefs than four Right Angles, which is abfurd. Therefore AB is not equal to LX. I fay alfo it is neither lefs than LX; for if this be poffible, make XO equal to AB, and XP to BC, and join OP. Then because A B is equal to BC, XO fhall be equal to XP; and the remaining Part OL equal to the remaining Part PM: And fo LM is † parallel to OP, and the † 2.6 Triangle LMX is equiangular to the Triangle OPX. Wherefore XL is ‡ to LM, as XO is to OP; and ‡ 4. 6, (by Alternation) as XL is to XO, fo is LM to OP. But LX is greater than XO. Therefore LM fhall also be greater than OP. But LM is put equal to A C. Wherefore, AC fhall be greater than OP. And so because the two Right Lines AB, BC, are equal to the two Right Lines OX, XP, and the Base AC greater than the Base OP; the Angle ABC will be greater than the Angle OXP. In like 25. I. manner, we demonftrate that the Angle DEF is greater than the Angle MXN, and the Angle GHK, than the Angle NXL. Therefore the three Angles ABC, DEF, GHK, are greater than the three Angles LXM, MXN, NXL. But the Angles ABC, DEF, GHK, are put lefs than four Right Angles. Therefore the Angles LX M, MXN, NX L, fhall be lefs by much than four Right Angles, and alfo equal to four Right Angles; which is abfurd, † Cor. 15. 1. Wherefore AB is not less than LX. It has also been prov'd not to be equal to it. Therefore it must neceffarily be greater. On the Point X raise † XR, ‡ 12 of this. perpendicular to the Plane of the Circle LMN; whofe Length let be fuch, that the Square thereof be equal to the Excefs, by which the Square of AB ex- / ceeds the Square of LX; and let RL, RM, RN, be joined. Because RX is perpendicular to the Plane of the Circle LMN, it fhall also be * perpendicular* Def. 3. to LX, MX, NX. And because LX is equal to * * 4. I. † 47. I. 18. 1. * 20. I. XM, and XR is common, and at Right Angles to them, the Bafe LR fhall be * equal to the Bafe RM. For the fame Reafon, RN is equal to RL, or RM. Therefore three Right Lines, RL, RM, RN, are equal to each other. And because the Square of X R is equal to the Excefs, by which the Square of AB exceeds the Square of LX; the Square of AB will be equal to the Squares of LX, XR together: But the Square of RL is equal to the Squares of LX, XR: For LXR is a Right Angle. Therefore the Square of AB will be equal to the Square of RL; and fo AB is equal to RL. But BC, DE, EF, GH, HK, are every of them equal to AB; and RN, or RM, equal to RL. Wherefore AB, BC, DE, EF, GH, HK, are each equal to RL, RM, or RN: And fince the two Sides RL, RM, are equal to the two Sides AB, BC, and the Bafe LM is put equal to the Bafe AC, the Angle LRM fhall be equal to the Angle ABC. For the fame Reafon the Angle MRN is equal to the Angle DEF, and the Angle LRN equal to the Angle GHK. Therefore a folid Angle is made at R of three plane Angles LRM, MRN, LRN, equal to three plane Angles given, ABC, DEF, GHK. * Now let the Center of the Circle X be in one Side of the Triangle, viz. in the Side MN, and join XL. I fay again, that AB is greater than LX. For if it be not fo, AB will be either equal, or lefs than LX. Firft let it be equal, then the two Sides AB, BC, are equal to the two Sides MX, LX, that is, they are equal to MN; but MN is put equal to DF. Therefore DE, EF, are equal to DF, which is impoffible. Therefore AB is not equal to LX. In like manner, we prove that it is neither leffer; for the Abfurdity will much more evidently follow. Therefore AB is greater than LX. And if in like manner, as before, the Square of RX be made equal to the Excefs, by which the Square of AB exceeds the Square of LX, and RX be raised at Right Angles to the Plane of the Circle, the Problem will be done. Laftly, let the Center X of the Circle be without the Triangle LMN, and join LX, MX, NX. I fay AB is greater than LX. For if it be not, it muft muft either be equal, or lefs. First, let it be equal; then the two Sides AB, BC, are equal to the two Sides MX, XL, each to each; and the Bafe AC is equal to the Base ML; therefore the Angle ABC is equal to the Angle MXL. For the fame Reason, the Angle GHK is equal to the Angle LXN; and fo the whole Angle MXN is equal to the two Angles ABC, GHK; but the Angles ABC, GHK, are greater than the Angle DĚF. Therefore the 2. Angle MXN is greater than DEF; but because the two Sides DE, EF, are equal to the two Sides MX, XN, and the Base DF is equal to the Base MN, the Angle MXN shall be equal to the Angle DEF; but it has been proved greater, which is abfurd. Therefore AB is not equal to LX. Moreover we will prove that it is not lefs; wherefore it shall be necef farily greater. And if, again, XR be raised at Right Angles to the Plane of the Circle, and made equal to the Side of that Square, by which the Square of A B exceeds the Square of LX, the Problem will be determined. Now, I fay, AB is not lefs than LX; for if it is poffible that it can be lefs, make XO equal to AB, and XP equal to B C, and join OP. Then, because AB is equal to BC, XO fhall be equal to XP, and the remaining Part OL equal to the remaining Part PM; therefore LM is parallel to PO, and the Triangle LMX equiangular to the Triangle PXO. Wherefore as + XL is to LM, fot4.6. is XO to OP: And (by Alternation) as L X is to XO, so is LM to OP; but LX is greater than XO'; therefore LM is greater than OP; but LM is equal to AC; wherefore AC fhall be greater than OP. And so because the two Sides AB, BC, are equal to the two Sides OX, XP, each to each; and the Base AC is greater than the Bafe OP; the Angle ABC fhall be greater than the Angle OX P. So likewife ‡ 25. if XR be taken equal to XO or XP, and OR be joined, we prove that the Angle GHK is greater than the Angle OXR. At the Point X, with thể Right Line LX, make the Angle LXS equal to the Angle ABC, and the Angle LXT equal to the Angle GHK, and XS, XT, each equal to XO, and join OS, OT, ST. Then becaufe the two Sides AB, BC, are equal to the two Sides OX, XS, and the Angle ABC is equal to the Angle OXS, the Bafe AC; that is, LM fhall be equal to the Base OS. For the fame Reason, LN is alfo equal to OT. And fince the two Sides ML, LN, are equal to the two Sides OS, OT, and the Angle MLN greater than the Angle SOT; the Base MN fhall be greater than the Bafe ST; but MN is equal to DF; therefore DF fhall be greater than ST. Wherefore because the two Sides DE, EF, are equal to the two Sides SX, XT, and the Bafe DF is greater than the Bafe ST, the Angle DEF fhall be greater than the Angle SXT; but the Angle SXT is equal to the Angles ABC, GHK. Therefore the Angle DEF, is greater than the Angles ABC, GHK; but it is alfo lefs, which is abfurd; which was to be de monftrated. PROPOSITION XXIV. THEOREM. If a Solid be contained under fix parallel Planes, the oppofite Planes thereof, are equal Parallelograms. ET the Solid CDGH be contained under parallel Planes AC, GF, BG, CE, FB, AE. I fay, the oppofite Planes thereof are equal Parallelograms. For because the parallel Planes BG, CE, are cut * 16 of this. by the Plane AC, their common Sections are * parallel; wherefore AB is parallel to CD. Again, because the two parallel Planes BF, AE, are cut by the Plane A C, their common Sections are parallel; therefore AD is parallel to BC; but AB has been proved to be parallel to CD; wherefore AC fhall be a Parallelogram. After the fame Manner, we demonftrate that CE, F G, GB, BF, or AE, is a Parallelogram. Let AH, DF, be joined. Then because AB is parallel to DC, and BH to CF, the Lines AB, BH, touching each other, fhall be parallel to the Lines DC, CF; touching each other, and not being in the +10 of this. fame Plane; wherefore they fhall + contain equal Angles. And fo the Angle ABH is equal to the_Angle DCF. DCF. And fince the two Sides AB, BH, are ‡ e- ‡ 34. I. qual to the two Sides D C, CF, and the Angle ABH equal to the Angle DCF, the Bafe AH fhall be** 4.1. equal to the Bafe DF, and the Triangle ABH equal to the Triangle DCF. And fince the Parallelogram BG is † double to the Triangle ABH, and the Pa- † 41. 1. rallelogram CE, to the Triangle DCF, the Parallelogram BG fhall be equal to the Parallelogram CE. In like manner, we demonftrate that the Parallelogram AC is equal to the Parallelogram GF, and the Parallelogram AE equal to the Parallelogram BF. If, therefore, a Solid be contained under fix parallel Planes, the oppofite Planes thereof are equal Parallelograms; which was to be demonstrated. Coroll. It follows from what has been now demonftrated, that if a Solid be contained under fix parallel Planes, the oppofite Planes thereof are fimilar and equal, because each of the Angles are equal, and the Sides about the equal Angles are proportional. PROPOSITION XXV. THE ORE M. If a folid Parallelepipedon be cut by a Plane, parallel to oppofite Planes; then as Bafe is to Base, fo fhall Solid be to Solid. TET the folid Parallelepipedon ABCD, be cut L by a Plane Y E, parallel to the oppofite Planes RA, DH. I fay as the Bafe EFA is to the Base For let AH be both Ways produced, and make P 3 |