Sidebilder
PDF
ePub

therefore as the folid AB to the folid CV, fo is the folid CD to Book XI. the folid CV; that is, each of the folids AB, CD has the fame ratio to the folid CV; and therefore the folid AB is equal to the folid CD.

Second general Cafe. Let the infisting straight lines FE, BL, GA, KH; XN, DO, MC, RP not be at right angles to the bases of the solids; and from the points F, B, K, G; X, D, R, M draw perpendiculars to the planes in which are the bases EH, NP, meeting those planes in the points S, Y, V, T; Q, I, U, Z; and complete the folids FV, XU, which are parallelepipeds, as was proved in the last part of Prop. 31. of this Book. In this cafe likewife, if the folids AB, CD be equal, their bases are reciprocally proportional to their altitudes, viz. the base EH to the base NP, as the altitude of the folid CD to the altitude of the folid AB. Because the folid AB is equal to the folid CD, and that the folid BT

g

is equal to the folid BA, for they are upon the fame base FK, g. 29.013@, and of the fame altitude; and that the folid DC is equal to the

II.

[blocks in formation]

folid DZ, being upon the fame base XR, and of the same altitude; therefore the folid BT is equal to the folid DZ. but the bases are reciprocally proportional to the altitudes of equal folid parallelepipeds of which the infisting straight lines are at right angles to their bafes, as before was proved. therefore as the base FK to the base XR, fo is the altitude of the folid DZ to the altitude of the folid BT. and the base FK is equal to the bafe EH, and the base XR to the base NP. wherefore as the bafe EH to the bafe NP, fo is the altitude of the folid DZ to the altitude of the folid BT. but the altitudes of the folids DZ, DC, as alfo of the folids BT, BA are the fame. Therefore as the base EH to the base NP, fo is the

Book XI. altitude of the folid CD to the altitude of the folid AB; that is, the bafes of the folid parallelepipeds AB, CD are reciprocally proportional to their altitudes.

Next, Let the bases of the folids AB, CD be reciprocally proportional to their altitudes, viz. the base EH to the bafe NP, as the altitude of the folid CD to the altitude of the folid AB; the folid AB is equal to the folid CD. the fame construction being made; because as the bafe EH to the bafe NP, fo is the altitude of the folid CD to the altitude of the folid AB; and that the base EH is equal to the bafe FK; and NP to XR; therefore the base FK is to the base XR, as the altitude of the folid CD to the

[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][subsumed][ocr errors][ocr errors][merged small][subsumed][merged small][merged small][subsumed]

altitude of AB. but the altitudes of the folids AB, BT are the fame, as also of CD and DZ; therefore as the base FK to the base XR, fo is the altitude of the folid DZ to the altitude of the folid BT. wherefore the bafes of the folids BT, DZ are reciprocally proportional to their altitudes; and their insisting straight lines are at right angles to the bafes; wherefore, as was before proved, the g. 29.or 30. folid BT is equal to the folid DZ. but BT is equal to the folid BA, and DZ to the folid DC, because they are upon the fame bafes, and of the fame altitude. Therefore the folid AB is equal to the folid CD. Q. E. D.

II.

PROP. XXXV. THEOR.

Book XI.

F from the vertices of two equal plane angles there See N.

It

be drawn two straight lines elevated above the planes in which the angles are, and containing equal angles with the fides of those angles, each to each; and if in the lines above the planes there be taken any points, and from them perpendiculars be drawn to the planes in which the first named angles are; and from the points in which they meet the planes, ftraight lines be drawn to the vertices of the angles first named; these straight lines fhall contain equal angles with the ftraight lines which are above the planes of the angles.

Let BAC, EDF be two equal plane angles; and from the points A, D let the straight lines AG, DM be elevated above the planes of the angles, making equal angles with their fides, each to each; viz. the angle GAB equal to the angle MDE, and GAC to MDF;

TAN

[blocks in formation]

and in AG, DM let any points G, M be taken, and from them let perpendiculars GL, MN be drawn to the planes BAC, EDF meeting these planes in the points L, N; and join LA, ND. the angle GAL is equal to the angle MDN.

Make AH equal to DM, and thro' H draw HK parallel to GL. but GL is perpendicular to the plane BAC, wherefore HK is perpendicular a to the fame plane. from the points K, N, to the a. s. II. ftraight lines AB, AC, DE, DF, draw perpendiculars KB, KC, NE, NF; and join HB, BC, ME, EF. because HK is perpendicular

b. 18. II.

Book XI. to the plane BAC, the plane HBK which paffes through HK is at right angles to the plane BAC; and AB is drawn in the plane BAC at right angles to the common section BK of the two planes; c.4.Def.11. therefore AB is perpendicular to the plane HBK, and makes right d.3.Def.11. angles d with every straight line meeting it in that plane. but BH

meets it in that plane; therefore ABH is a right angle. for the fame reason, DEM is a right angle, and is therefore equal to the angle ABH. and the angle HAB is equal to the angle MDE. therefore in the two triangles HAB, MDE there are two angles in one equal to two angles in the other, each to each, and one fide equal to one fide, oppofite to one of the equal angles in each, viz. HA e. 26. 1. equal to DM; therefore the remaining fides are equal ✨, each to each. wherefore AB is equal to DE. In the fame manner, if HC and MF be joined, it may be demonstrated that AC is equal to DF. therefore fince AB is equal to DE, BA and AC are equal

[blocks in formation]

f. 4. I.

H

G

to ED and DF; and the angle BAC is equal to the angle EDF; wherefore the bafe BC is equal to the bafe EF, and the remaining angles to the remaining angles. the angle ABC is therefore equal to the angle DEF. and the right angle ABK is equal to the right angle DEN, whence the remaining angle CBK is equal to the remaining angle FEN. for the fame reason, the angle BCK is equal to the angle EFN. therefore in the two triangles BCK, EFN there are two angles in one equal to two angles in the other, each to each, and one fide equal to one fide adjacent to the equal angles in each, viz. BC equal to EF; the other fides therefore are equal to the other fides; bK then is equal to EN. and AB is equal to DE; wherefore AB, BK are equal to DE, EN; and they contain right angles; wherefore the base AK is equal to the bafe DN. and fince AH is equal to DM, the fquare of AH is equal to the

fquare of DM. but the fquares of AK, KH are equal to the fquare Book XI. 8 of AH, because AKH is a right angle. and the squares of DN, NM are equal to the fquare of DM, for DNM is a right angle. g. 47. I. wherefore the squares of AK, KH are equal to the squares of DN, NM; and of those the square of AK is equal to the square of DN. therefore the remaining square of KH is equal to the remaining fquare of NM; and the ftraight line KH to the straight line NM. and because HA, AK are equal to MD, DN, each to each, and the base HK to the bafe MN, as has been proved; therefore the angle HAK is equal to the angle MDN. Q. E. D.

COR. From this it is manifeft, that if from the vertices of two equal plane angles there be elevated two equal straight lines containing equal angles with the fides of the angles, each to each; the perpendiculars drawn from the extremities of the equal straight lines to the planes of the first angles are equal to one another.

Another Demonstration of the Corollary.

Let the plane angles BAC, EDF be equal to one another, and let AH, DM be two equal straight lines above the planes of the angles, containing equal angles with BA, AC, ED, DF, each to each, viz. the angle HAB equal to MDE, and HAC equal to the angle MDF; and from H, M let HK, MN be perpendiculars to the planes BAC, EDF; HK is equal to MN.

h. 8. 1.

Because the folid angle at A is contained by the three plane angles BAC, BAH, HAC, which are, each to each, equal to the three plane angles EDF, EDM, MDF containing the folid angle at D; the folid angles at A and D are equal, and therefore coincide with one another; to wit, if the plane angle ABC be applied to the plane angle EDF, the straight line AH coincides with DM, as was fhewn in Prop. B. of this Book. and because AH is equal to DM, the point H coincides with the point M. wherefore HK which is perpendicular to the plane BAC coincides with MN i. 13. 11. which is perpendicular to the plane EDF, because these planes coincide with one another. therefore HK is equal to MN. Q. E. D.

i

« ForrigeFortsett »