1 Make A H equal to DM, and through H let HK be drawn parallel to GL; but GL is perpendicular to the Plane paffing thro BAC, therefore KH fhall be + + 8 of biso allo perpendicular to the Plane paffing through BAC: Draw from the Points K and N, to the Right Lines AB, AC, DE, and DF, the Perpendiculars KB, KC, NE, NF; and join H C, CB, MF, FE: Then, be caufe the Square of HA is equal to the Squares of 1 47. 1. HK, KA, and the Squares of K C and C A are † equal to the Square of KA; the Square of H A fhall be equal to the Squares of H K, KC, and CA: But the Square of HC is equal to the Squares of H K and KC; therefore the Square of HA will be equal to the Squares of HC and CA; and fo the Angle HCA is † † 48. 1. a Right Angle. For the fame Reafon, the Angle DFM is alfo a Right Angle; therefore the Angle ACH is equal to DFM: But the Angle HAC is alfo equal to the Angle MDF; therefore the two Triangles MDF, HAC, have two Angles of the one equal to two Angles of the other, each to each, and one Side of the one equal to one Side of the other; viz. that which is fubtended by one of the equal Angles: that is, the Side HA equal to DM; and fo the other Sides of the one fhall be equal to the other Sides of the other, * 26. 1. each to each: Wherefore A C is equal to DF. In like manner we demonftrate, that A B is equal to DE: For, let A B, ME, be joined; then, becaufe the Square of AH is equal to the Squares of A K and KH; and the Squares of A Band BK are. equal to the Square of A K; the Squares of A B, BK, and K H, will be equal to the Square of A H. But the Square of BH is equal to the Squares of B K, KH; for the Angle H K B is a Right Angle, because HK is perpendicular to the Plane paffing thro' B AC; therefore the Square of A H is equal to the Squares of AB and BH: Wherefore the Angle A B His† at 48, 1. Right Angle. For the fame Reafon the Angle DEM is alfo a Right Angle; and the Angle BAH is equal to the Angle EDM, for fo it is put; and A His equal to DM; therefore A B is * alfo equal to DE: * And fo, fince AC is equal to DF, and A B to DE; the two Sides CA, A B, shall be equal to the two Sides FD, DE: But the Angle B A C is equal to the Angle FDE; therefore the Bafe BC is equal to * Q.4 * the 4. * the Bafe EF, the Triangle to the Triangle, and the other Angles to the other Angles: Wherefore the Angle ACB is equal to the Angle DF E. But the Right Angle ACK is equal to the Right Angle D F Ñ; and therefore the remaining Angle BCK is equal to the remaining Angle EFN. For the fame Reafon, the Angle CBK is equal to the Angle FEN; and fo, because BC K and EFN are two Triangles, having two Angles equal to two Angles, each to each, and one Side equal to one Side, which is at the equal Angles; viz. BC equal to E F; therefore they fhall have the other Sides equal to the other Sides: Therefore C K is equal to FN. But AC is equal to DF; therefore the two Sides A C, CK, are equal to the two Sides DF, FN; and they contain Right Angles; confequently, the Bafe AK is equal to the Bafe D N. And fince A H is equal to DM, the Square of A H fhall be equal to the Square of D M; But the Squares of AK and KH are equal to the Square of A H; for the Angle A KH is a Right Angle; and the Squares of D N and N M ; of which the Square of A K is equal to the Square of DN: Wherefore the Square of K H remaining is equal to the remaining Square of NM; and fo the Right Line H K is equal to M N. And fince the two Sides HA, A K, are equal to the two Sides M D, DN, each to each, and the Bafe K H has been proved equal to the Base N M, the Angle H A K, or GA L, fhall be + equal to the Angle MDN; which was to be demonftrated. Coroll. From hence it is manifeft, that if there be two Right-lined plane Angles equal, from whofe Points equal Right Lines be elevated on the Planes of the Angles, containing equal Angles with the Lines first given, each to each; Perpendiculars drawn from the extreme Points of thofe elevated Lines to the Planes of the Angles firft given, are equal to one another. PRO i PROPOSITION XXXVI. THEOREM. If three Right Lines be proportional, the folid Pa rallelopipedon made of them is equal to the folid Parallelopipedon made of the middle Line, if it be an equilateral one, and equiangular to the aforefaid Parallelopipedon. L ET three Right Lines A, B, C, be proportional; viz. let A be to B, as B is to C. I fay, the Solid made of A, B, C, is equal to the equilateral Solid made of B, equiangular to that made on A, B, C. Let E be a folid Angle contained under the three plane Angles DEG, GEF, FED; and make DE, GE, EF, each equal to B, and compleat the folid Parallelopipedon EK: Again, put L M equal to A, and at the Point L, at the Right Line L M, make a folid 26 of th Angle contained under the plane Angles NLX, XLM, MLN, equal to the folid Angle E; and make LN equal to B, and L X equal to C: Then, because A is to B, as B is to C; and A is equal to LM; and B to LN, EF, EG, or ED; and C to LX; it fhall be, as LM is to EF, fo is GE to LX: And fo the Sides about the equal Angles MLX, GEF, are reciprocally proportional. Wherefore the Parallelogram MX + is † 14.6x equal to the Parallelogram GF. And fince the two plane Angles GEF, XLM, are equal, and the Right Lines LN, ED, being equal, are erected at the angular Points containing equal Angles with the Lines first given each to each; the Perpendiculars drawn ‡ from Cor. 35. of the Points N and D, to the Planes drawn thro' XLM, this. GEF, are equal one to another: Therefore the Solids LH, EK, have the fame Altitude. But folid Parallelopipedons that have equal Bafes, and the fame Altitude, are equal to each other; therefore the Solid 31 of this, HL is equal to the Solid E K. But the Solid H L is that made of the three Right Lines A, B, C, and the Solid EK, that made of the Right Line B: Therefore, if three Right Lines be proportional, the folid Parallelopipedon made of them is equal to the folid Parallelopipedon made of the middle Line, if it be an equilateral one, * 33 of this. and equiangular to the aforefaid Parallelopipedon; which was to be demonftrated. If four Right Lines be proportional, the folid Parallelopipedons fimilar, and in like manner defcribed from them, shall be proportional. And if the faid Parallelopipedons, being fimilar, and alike defcribed, be proportional, then the Right Lines they are defcribed from, fhall be propor tional. LET the four Right Lines AB, CD, EF, GH, be proportional; viz let A B be to CD, as E F is to GH; and let the fimilar and alike fituate Parallelopi pedons KA, LC, ME, N G, be defcribed from them. I fay, KA is to L C, as M E is to N G. For, because the folid Parallelopipedon K.A is fimilar to L C, therefore K A to LC fhall be a Proportion triplicate of that which A B has to CD. For the fame Reafon, the Solid ME to NG will have a triplicate Proportion of that which L F has to G H. But AB is to CD, as EF is to GH; therefore A K is to LC, as ME is to.N G. And if the Solid AK be to the Solid LC, as the Solid ME is to the Solid N G; I fay, as the Right Line A B is to the Right Line CD, fo is the Right Line E F to the Right Line GH: For, +33 of this. because A K to LC has + a Proportion triplicate of that which A B has to CD; and ME to G N has a Proportion triplicate of that which E F has to GH; and fince AK is to LC, as ME is to NG; it fhall be, as AB is to CD, 'fo is EF to GH. Therefore, if four Right Lines be proportional, the folid Parallelopipedons 'fimilar, and in like manner defcribed from them, fhall be proportional. And if the folid Parallelopipedons, being fimilar and alike described, be proportional, then the Right Lines they are defcribed from, shall be proportional; which was to be demonftrated. PRO PROPOSITION XXXVIII. THEOREM. If a Plane be perpendicular to a Plane, and a LET the Plane CD be perpendicular to the Plane * per this. this. For, if it does not, let it fall without the fame, as EF, meeting the Plane A B in the Point F; and from the Point F let FG be drawn from the Piane A B, perpendicular to A D; this fhall be perpendicular to the Def. 4. of Plane CD; and join EG: Then, because FG is pendicular to the Plane CD, and the Right Line EG, in the Plane CD, touches it; the Angle FGE fhall be + a Right Angle. But E F is alfo at Right Angles + Def. 3. of to the Plane AB; therefore the Angle EFG is a Right Angle: And fo, two Angles of the Triangle EFG are equal to two Right Angles; which is ab- ‡ 17 furd. Wherefore, a Right Line drawn from the Point E perpendicular to the Plane A B, does not fall without the Right Line A D; and fo it must neceffarily fall on it. Therefore, if a Plane be perpendicular to a Plane, and a Line be drawn from a Point in one of the Planes perpendicular to the other Plane; that Perpendicular fhall fall in the common Section of the Planes; which was to be demonftrated, PRO |