XM, and XR is common, and at Right Angles to them, the Base LR shall be * equal to the Base RM. For the same Reason, RN is equal to RL, or RM. Therefore three Right Lines, RL, RM, RN, are equal to each other. And because the Square of XR is equal to the Excess, 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 tequal to the Squares of LX, XR: For LXR is a Right Angle. Therefore the Square of A B will be equal to the Square of RL; and fo AB is equal to RL. But B C, 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 since the two Sides RL, RM, are equal to the two Sides AB, BC, and the Base LM is put equal to the Base AC, the Angle LRM shall be f equal to the Angle ABC. For the fame Reason the Angle MRN is equal to the Angle DEF, and the Angle LRN equal to the Angle GHK. Therefore a solid 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 so, AB will be either equal, or less than LX. First let it be equal, then the two Sides A B, 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 * impossible. Therefore AB is not equal to LX. In like manner, we prove that it is neither lesser ; 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 Excess, by which the Square of AB exceeds the Square of L X, and RX be raised at Right Angles to the Plane of the Circle, the Problem will be done. Lastly, 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 20. I. must either be equal, or less. First, let it be equal ; then the two Sides AB, BC, are equal to the two Sides MX, XL, each to each; and the Base AC is equal to the Base ML; therefore the Angle A B C is equal to the Angle MXL. For the fame Reason, the Angle GHK is equal to the Angle L XN; and fo the whole Angle MXN is equal to the two Angles ABC, GHK; but the Angles A B C, GHK, are greater than the Angle DĚF. Therefore the 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 L X. Moreover we will prove that it is not less; wherefore it shall be necessarily 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 AB exceeds the Square of LX, the Problem will be determined. Now, I say, AB is not less than LX; for if it is possible that it can be less, make X O equal to A B, and XP equal to B C, and join OP. Then, because AB is equal to BC, XO shall be equal to XP, and the remaining Part OL equal to the remaining Part PM; therefore LM is * parallel to * 2. 6; PO, and the Triangle LMX equiangular to the Triangle PXO. Wherefore as + XL is to LM, sot 4.6. 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 L M is greater than OP; but LM is equal to AC; wherefore AC shall 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 Base OP; the Angle ABC shall be I greater than the Angle O XP. So likewise 1 25.5' if XR be taken equal to X or XP, and OR be joined, we prove that the Angle GHK is greater than the Angle OXR. At the Point X, with the Right Line LX, make the Angle LXS equal to the Angle A B C, and the Angle LXT equal to the Angle GHK, and XS, XŤ, each equal to XO, and join OS, OT, ST. Then becaufe the two Sides AB, BC, are equal to the two Sides O X, XS, and P 2 the the Angle ABC is equal to the Angle OXS, the Base AC; that is, LM shall be equal to the Base OS. For the fame Reason, LN is also equal to OT. And since 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 Ihall be greater than the Base $T; but MN is equal to DF; therefore DF shall be greater than ST. Wherefore because the two Sides DE, EF, are equal to the two Sides SX, XT, and the Base DF is greater than the Base 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 also less, which is absurd ; which was to be done monstrated. PROPOSITION XXIV. THEOREM. the opposite Planes thereof, are equal Parallelo- * LET the Solid CP G H be contained under paralle GF, , . , the opposite Planes thereof are equal Parallelograms. For because the parallel Planes BG, CĒ, are cut 16 of tbis. by the Plane AC, their common Sections are paral lel ; 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 shall be a Parallelogram. After the fame Mannerwe demonstrate that CE, FG, GB, BF, or AE, is a Parallelogram. Let AH, DF, be joined. Then because AB is parallel to DC, and B H to CF, the Lines AB, BH, touching each other, shall be parallel to the Lines DC, CF; touching each other, and not being in the + 10 of this. fame Plane; wherefore they shall + contain equal Angles. And so the Angle A BH is equal to the Angle DCF. DCF. And since the two Sides AB, BH, are fe- 1 34. 1. Coroll. It follows from what has been now demon strated, that if a Solid be contained under fix paral- PROPOSITION XXV. THEOREM. rallel to opposite Planes ; tben as Base is to Base, L ET the solid Parallelepipedon ABCD, be cut by a Plane Y E, parallel to the opposite Planes RA, DH, I say as the Base E FOA is to the Base EHCF, so is the Solid ABFY to the Solid EGCD. For let AH be both Ways produced, and make HM, MN, &c, equal to E H, and AK, KL, &c. equal to AE; and let the Parallelograms LO, K4, HX, MS, as likewise the Solids LP, KR, H., MT, be compleated. Then because the Right Lines LK, KA, AE, are equal, the Parallelograms LO, Ko, AF, shall be * also equal; as likewise the Pa- * 1. 6. rallelograms KE, KB, AG: And moreover + the + 24 of tbis. Parallelograms Lt, KP, AR, for they are opposite, For the same Reason, the Parallelograms E C, HX, MS, also are equal to each other; as also the Paral P 3 lelograms lelograms HG, HI, IN; and so are the Parallelo grams DH, M4, NT. Therefore three Planes of the Solid LP, are equal to three Planes of the Solid KR, or AY, each to each; and the Planes oppofite to these, are equal to them. Therefore the three Solids 1 Def. 10. of LP, KR, AY, will be equal to each other. For the fame Reafon, the three Solids ED, HP, MT, are equal to each other. Therefore the Base LF is the same Multiple of the Base AF, as the Solid LY is of the Solid AY. For the fame Reason, the Base NF is the fame Multiple of the Base HF, as the Solid NY is of the Solid ED: And if the Base LF be equal to the Bafe NF, the Solid L Y shall be equal to the Solid NY; and if the Base LF exceeds the Base NF, the Solid LY shall exceed the Solid NY; and if it be less, less. Wherefore because there are four Magnitudes, viz. the two Bases AF, FH, and the two Solids AY, ED, whose Equimultiples are taken, to wit, the Base LF, and the Solid LY; and the Base NF, and the Solid NY: And since it is proved, if the Base LF exceeds the Base NF, then the Solid LY will exceed the Solid NY, if equal, equal, and less, less. Therefore as the Base AF is * D&f.6. 5. to the Base FH, fo is * the Solid AY to the Solid ED. Wherefore, if a solid Parallelepipedon be cut by a Plane, parallel to opposite Planes ; then as Bafe is to Bale, o fhall Solid be to Solid; which was to be demonstrated. PROPOSITION XXVI. THEO R E M. At a Right Line given, and at a Point given in it, to make a solid Angle equal to a folid Angle in it, and D a given folid Angle contained under the plane Angles EDC, EDF, FDC; it is required to make a solid Angle at the given Point A, in the given Right Line AB, equal to the given solid Angle D. Afiume any Point F in the Right Line DF, from * 11 of tbis. which let F Ġ be drawn * perpendicular to the Plane passing |