1 , Make A H equal to DM, and through H let HK bę drawn parallel to GL; but G L is perpendicular to the Plane paffing' thro? BAC ; therefore KH shall be † + 8 of ibis. also perpendicular to the Plane passing through BAC: Draw from the Points Kand N, to the Right Lines NE, NF; and join HC,CB, MF, FE: Then, be cause the Square of HA is I equal to the Squares of I 47. 1. H K, KA, and the Squares of K C and C A are + equal to the Square of K A; the Square of H A shall be equal to the Squares of HK, KC, and CA: But the Square of HC is equal to the Squares of H K and KC; therefore the Square of H A will be equal to the Squares of HC and CA ; and so the Angle HCA is + + 48. 1. a Right Angle. For the fame Reason, the Angle DFM is also a Right Angle; therefore the Angle A CH is equal to DFM: But the Angle HAC is also equal to the Angle MDF; therefore the two Triangles MDF, HAC, have two Angles of the-one equal to two Angies 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 H A equal to DM; and so the other Sides of the one shall 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 demonstrate, that A B is equal to DE: l'or, let A B, ME, be joined; then, because the Square of AH is equal to the Squares of A K and KH; and the Squares of A B and B K are, equal to the Square of A K; the Squares of A B, B'K, and K H, will be equal to the Square of AH. But the Square of B H 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 passing thro’BAC; therefore the Square of A H is equal to the Squares of A B and BH: Wherefore the Angle A BH is t a + 48. 1. Right Angle. For the same Reason the Angle DEM is also a Right Angle; and the Angle BA H is equal to the Angle EDM, for to it is put; and A H is equal to D M; therefore A B is * also equal to DE: * 4.1, 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 BAC is equal to the Angle FDE; therefore the Base B C is * equal o Q: 4 the 1 26. 1, the Base E F, the Triangle to the Triangle, and the other Angles to the other Angles: Wherefore the Ane gle ACB is equal to the Angle D F E. But the Righệ Angle AC K is equal to the Right Angle DFN; and therefore the remaining Angle BC K is equal to the remaining Angle EFN. For the same Reason, the Angle CBK is equal to the Angle F EN; and fo, because BC K and 'E FN are two Triangles, hay ing two Angles equal to two Angles, each to each, and one side equal to one side, which is at the equal Angles; viz. B C equal to E F; therefore * they shall have the other Sides equatro 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 ; consequently, * the Base A K is' equal to the Base D N. And since A H is equal to DM, the Square of A H shall 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 K H 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 N M; and so the Right Line HK is equal to MN, And since the iwo Sides HA, A K, are equal to the two Sides MD, DN, each to each, and the Bale K H has been proved equal to the Base N M, the Angle H AK, or GAL, shall be + equal to the Angle M DN; which was to be demonstrated, Coroll. From hence it is manifeft, that if there be two Right-lined plane Angles equal, from whose 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 those elevated Lines to the Planes of the Angles first given, are equal to one another. + 8.1. PRO PROPOSITION XXXVI... THEOREM. If tbree Right Lines be proportional, the solid Pa. rallelopipedon made of them is equal to the solid Parallelopipedon made of the middle Line, if it be an equilateral one, and equiangular to the aforesaid Parallelopipedon. LET three Right Lines A, B, C, be proportiona! . BB is I 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 solid Angle contained under the three plane Angles DEG, G EF, FED; and make DE, GE, EF, each equal to B, and compleat the solid Pa. rallelopipedon EK: Again, put L M equal to A, and at the Point L, at the Right Line L M, make * a solid • 26 of 10's, Angle contained under the plane Angles NL,X, XLM, ML N, equal to the solid Angle E; and make LN equal to B, and LX 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 Cto LX; it shall be, as LM is to EF, lo is GE to LX: And so the sides about the equal Angles M L X, GEF, are reciprocaliy proportional. Wherefore the Parallelogram MX † is † 14. Gay equal to the Parallelogram GF. And fince the two plane Angles G EF, XLM, are equal, and the Right Lines LN, ED, being equal, are erected at the angular Points containing equal Angles with the Lines hift given each to each; the Perpendiculars drawn from I Cor. 35. of the Points N and D, to the Planes drawn thro' XLM, Sbis. GEF, are equal one to another : Therefore the Solids LH, EK, have the same Altitude. But folid Paralle. lopipedons that have equal Bases, and the fame Altitude, are equal to each other; therefore the Solid • 31 of this, HL is equal to the Solid EK. 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 solid Parallelopipedon made of them is equal to the solid Parallelopipehon made of the middle Line, if it be an equilateral one, am * and equiangular to the aforesaid Parallelopīpedon ; which was to be demonstrated, PROPOSITION XXXVII. THEOREM. rallelopipedons similar, and in like manner de- 33 of this. LET the four Right Lines AB, CD, EF, GH, be proportional; viz let A B be to CD, as EF is to GH; and let the fimilar and alike fituare Parallelopi, pedons KA, LC, ME, NG, be defcribed from them. I say, K A is to LC, as M E is to N G. For, because the foliu Parallelopipedon KA is fimiJar to L C, therefore K A to LC shall be * a Proportion triplicate of that which AB has to CD. For the fame Reafon, the Solid M E to NG will have a triplicate Proportion of that which E F has te G H. But A B is to CD, as EF is to GH; therefixe a K is to LC, as M E is to.NG. And if the Solid AK be to the Solid LC, as the Solid ME is to the Solid NG;I fay, as the Right L ne A B is to the Right Line CD, fo is the Right Line E F to the Right Line GH: For, + 33 of tbis, because A K to L C has + a Proponion triplicate of that which A B bas to CD; and ME to G N has a Proportion triplicate of that which: E F has w GH; and fince AK is to LC, as ME is 10 NG it shall be, as AB is to CD, fo is EF to GH. Therefore, if four Right Lines be proportional, the folid Parallelepipedons fimilar, and in like manner described from tbem, shall be proportional. And if the folid Parallelopipedons, being imilar and alike described, be proportional, then the Right Lines they are described from, shall be proportional; which was to be demonstrated. PRO PROPOSITION XXXVIII. THBOR E M. If a Plane be perpendicular. to a Plant, and a Line be drawn from a Point in one of the of the Planes. A B, let their common Section be AD, and let fome Point E be taken in the Plane CD. I say, a Perpendicular drawn from the Point E to the Plane AB, falls on A D. For, if it does not, let it fall without the fame, as E F, meeting the Plane A B in the Point F; and from the Point Flet F G be drawn from the Plane A B, perpendieular to AD; this shall be * perpendicular to the * Def. 4. of Plane CD; and join EG: Then, because FG is perpendicular to the Plane CD, and the Right Line ĖG, in the Plane CD, touches it; the Angle F G E malli be + a Right Angle. But E F is also at Right Angles Def. 3. of 10 the Plane AB ; therefore the Angle E F G is a Right Angle: And so, two Angles of the Triangle EF G are equal to two Right Angles; which ist ab. 1 17.1 surd. 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 so it must neceffarily fall on it. Therefore, if a Plane be perpendicular to a Piane, and a Line be drawn from a Point in one of the Planes perpendicular to the other Plane ; that Perpendicular shall fall in the common Section of the Planes ; which was to be demonstrated, Ibis., 1 > tbis. PRO |