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XN, DO, MC, and RP be not at right angles to their bases. The base EH is to the base NP, as the altitude of CD is to the altitude of A B. From the points F, B, K, and G; X. D, R, and M, draw perpendiculars to the planes in which are the bases E H and N P, meeting those planes in the points S, Y, V, and T; Q, I, U, and Z; and complete the solids FV and X U, which are parallelopipeds (XI 31).

Because the solid A B is equal to the solid CD, and the solid AB is equal (XI. 29 or 30) to the solid BT, for they are upon the same base FK, and of the same altitude; and the solid CD is equal (XI. 29 or 30) to the solid DZ, being upon the same base XR, and of the same altitude. Therefore the solid BT is equal to the solid DZ. But the bases and altitudes of equal parallelopipeds of which the insisting straight lines are at right angles H to their bases, are reciprocally proportional. Therefore the base FK is to the base XR, as the altitude

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of the solid DZ is to the altitude of the solid BT: and the base FK is equal to the base EH, and the base X R to the base N P. Therefore, the base EH is to the base NP, as the altitude of the solid DZ is to the altitude of the solid BT: but the altitudes of the solids DZ and DC, as also of the solids B T and B A, are the same. Therefore the base EH is to the base N P, as the altitude of the solid CD is to the altitude of the solid A B; that is, the bases and altitudes of the parallelopipeds A B and CD are reciprocally proportional.

Conversely, let the bases of the solids A B and CD be reciprocally proportional to their altitudes, viz., the base EH to the base NP, as the altitude of the solid CD to the altitude of the solid AB. The solid AB shall be equal to the solid CD..

The same construction being made; because the base EH is to the base NP, as the altitude of the solid CD is to the altitude of the solid AB; and the base EH is equal to the base FK, and NP to XR. Therefore the base FK is to the base X R, as the altitude of the solid CD to the altitude of A B. But the altitudes of the solids A B and B T are the same, as also of CD and D Z. Therefore the base F K is to the base X R, as the altitude of the solid DZ is to the altitude of the solid BT. Wherefore the bases of the solids BT and DZ are reciprocally proportional to the altitudes: and their insisting straight lines are at right angles to their bases. Therefore, as was before proved, the solid BT is equal to the solid D Z. But BT is equal (XI. 29 or 30) to B A, and D Z to DC, because they are upon the same bases, and of the same altitude. Therefore the solid A B is equal to the solid CD. Therefore the bases, &c. Q. E. D.

PROP. XXXV. THEOREM.

If, from the vertices of two equal plane angles, there be drawn two straight lines elevated above the planes in which the angles are, and containing equal angles with the sides 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 those planes; and if from the points in which they meet the planes, straight lines be drawn to the vertices of the two equal plane angles; these straight lines make equal angles with the straight lines above the planes.

Let B A C and EDF be two equal plane angles; and from the points A and D let the straight lines A Ĝ and DM be drawn above the planes of the angles, making equal angles with their sides, viz., GAB equal to MDE, and G A C to MDF. From G and M, any points in the straight lines AG and DM, let perpendiculars GL and MN be drawn (XI. 11) to the planes BAC and EDF meeting them in the points L and N; and let AL and MN be joined. The angle GAL is equal to the angle MDN. Make A H equal to DM, and through H draw HK parallel to GL in the plane AGL, and meeting AL in K.

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M

H

D

Because GL is perpendicular to the plane BAC, H K is perpendicular to the same plane. Because the solid angle at A, is contained by three plane angles B AC, BAH, HA C, which are equal, each to each, to the three plane angles EDF, EDM, MDF, which contain the solid angle at D. Therefore the solid angles at A and D are equal, and coincide with one another (XI. B); that is, if the plane angle B A C be applied to the plane angle EDF, the straight line A H coincides with DM. Because A H is equal to DM, the point H coincides with the point M. Therefore HK, which is perpendicular to the plane BAC, coincides with MN (XI. 13) which is perpendicular to the plane ED F, and HK is equal to MN. Because the points A and K coincide with the points D and N, the straight line A K coincides with the straight line DN. Therefore the triangle AHK coincides with the triangle DMN, and the angle HAK with the angle MDN. Wherefore the angle GAL is equal to the angle MDN. Therefore, if from the vertices, &c. Q. E. D.

COROLLARY.-From this it is manifest, that if from the vertices of two equal plane angles, there be drawn two equal straight lines containing equal angles with the sides 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.

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If three straight lines be proportionals, the parallelopiped described from all three, as its sides, is equal to the equilateral parallelopiped described from the mean proportional, one of the solid angles of which is contained by three plane angles equal, each to each, to the three plane angles containing one of the solid angles of the other. Let A, B, C be three proportionals, viz., A to B, as B to C. The

parallelopiped described from A, B, C as its edges shall be equal to the equilateral parallelopiped described from B, as its common edge, equiangular to the other.

Take a solid angle D contained by three plane angles EDF, FDG, and GDE; and make each of the straight lines ED, DF, and DG equal to B, and complete the parallelopiped D H. Take any straight line LK equal to A, and at the point K in the straight line LK, make (XI. 26) a solid angle contained by the three plane angles L KM, MKN and NK L, equal to the angles EDF, FDG, and GD E, each to each; make KN equal to B, and KM equal to C; and complete the parallelopiped KO.

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Because, A is to B, as B is to C, and A is equal to LK, and B to each of the straight lines DE, D F, and C to K M. Therefore LK is to ED, as DE is to KM; that is, the sides about the equal angles are reciprocally proportional; therefore the parallelo gram LM is equal (VI. 14) to the parallelogram EF. Because EDF and LKM are two equal plane angles, and the two equal straight lines DG and K N are drawn from their vertices above their planes; and contain equal angles with their sides. Therefore the perpendiculars from the points G and N to the planes EDF and LK M are equal to (XI. 35 Cor.) one another. Wherefore the solids K O and DH are of the same altitude: and they are upon equal bases LM, and EF. Therefore the parallelopiped KO is equal (XI. 31) to the parallelopiped DH; and the solid KO is described from the three straight lines A, B, and C, as its edges, and the solid DH from the straight line B as its common edge. Therefore if three straight lines, &c. Q. E. D.

Exercise. Demonstrate the converse of this proposition.

PROP. XXXVII. THEOREM.

D

B

If four straight lines be proportionals, the similar parallelopipeds similarly described from them as edges are also proportionals; and, conversely, if the similar parallelopipeds similarly described from four straight lines as edges be proportionals, the straight lines are proportionals. Let the four straight lines A B, CD, EF and GH, be proportionals, viz., AB to CD, as EF to GH; and let the similar parallelopipeds AK, CL, E M and GN be similarly described from them as edges. AK is to CL, as EM is to GN.

Make (VI. 11) A B, CD, O and P continual proportionals, as also EF, GH, Q and R.

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Because A B is to CD, as EF to GH; and CD is (V. 11) to O, as GH is to Q; and O is to P, as Q is to R. Therefore, ex æquali (V. 22) AB is to P, as E F is to R. But A B is to P, as (XI. Cor. 33) the solid AK, is to the solid CL; and EF is to R, as (XI. Cor. 33) the solid EM is to the solid GN. There

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fore (V. 11) the solid AK is to the solid CL, as the solid E M is to the
solid GN.

Next, let the solid AK be to the solid
CL, as the solid EM to the solid G N.
The straight lines AB, CD, EF, and GH,
are proportionals.

M

N

F G H

Q

Take A B to CD, so EF to ST, and E from ST as edge, describe (XI. 27)a parallelopiped S V similar and similarly situated to either of the parallelopipeds, EM or G N.

Because A B is to CD, as E F is to ST, and that from A B and CD as edges the parallelopipeds AK and CL are similarly described; and the parallelopipeds EM and SV from the straight lines EF and ST as edges. Therefore AK is to CL, as EM is to S V. But (Hyp.) A K is to CL, as EM is to G N. Therefore G N is equal (V. 9) to SV. But it is similar and similarly situated to S V. Therefore the planes which contain the solids GN and SV are similar and equal, and their homologous sides G H and ST equal to one another. Because A B is to CD, as EF is to ST, and ST is equal to GH. Therefore A B is to CD, as EF is to GH. Therefore, if four straight lines, &c. Q. E. D.

PROP. XXXVIII. THEOREM.

If one plane be perpendicular to another, a straight line drawn from a point in the one perpendicular to the other, meets their common section.

Let the plane CD be perpendicular to the plane A B, and let AD be their common section; if any point E be taken in the plane CD, the perpendicular, drawn from E to the plane A B, meets A D.

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From E draw EG in the plane CD perpendicular to AD) (I. 12) and in the plane AB, from the point G, draw GF perpendicular to AD (I. 11).

Because EG is perpendicular to G F (XI. Def. 3), and also to AD (Const.) Therefore EG is perpendicular to the plane AB (XI. 4). But from the point E, no other straight line can be drawn perpendicular to A B, than EG (XI. 13). Therefore the perpendicular drawn from E to the plane AB meets AD. Wherefore, if one plane be perpendicular, &c. Q. E. D.

C

B

Dr. Simson considers this proposition as an interpolation in this Book, and quite out of place. :

PROP. XXXIX. THEOREM.

In a parallelopiped, if the sides of two of the opposite planes be each bisected, the common section of the planes passing through the points of bisection and any diagonal of the parallelopiped, bisect each other.

Let the sides of the opposite planes CF and A H, of the parallelopiped AF, be bisected in the points K, L, M, and N; and X, O, P, and R. Join KL, MN, XO, and PR. Because DK and CL are equal and parallel, KL is parallel (I. 33) to DC. For the same reason, M N is parallel to BA. But BA is parallel to DC. Because KL and BA are each of them parallel to DC, and not in the same plane with it, KL is parallel (XI. 9) to BA; and because KL and MN are each of them parallel to

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BA, and not in the same plane with it, KL is parallel (XI. 9) to MN. Therefore KL and MN are in one plane. In like manner it may be proved, that X O and PR are in one plane. Let YS be the common section of the planes KN and XR; and DG the diagonal of the parallelopiped A F. The straight lines Y S and DG meet and bisect each other. Join DY, YE, BS, and S G.

B

N

E

Because DX is parallel to OE, the alternate angles DXY and YOE are equal (I. 29). Because D X is equal to OE and X Y to Y Ó, and they contain equal angles, the base D Y is equal (I. 4) to the base YE, and the other angles are equal. Therefore the angle XYD is equal to the angle OŸE, and DYE is a straight (I. 14) line. For the same reason BSG is a straight line, and B S is equal to S G. Because CA is equal and parallel to DB, and also to EG. Therefore D B is equal and parallel (XI. 9) to EG; and DE and BG join their extremities. Therefore DE is equal and parallel (I. 33) to BG. But DG and Y S drawn joining them are in one plane. Therefore DG and YS must meet one another; let them meet in T. Because DE is parallel to B G, the alternate angles EDT and BGT are (XI. 29) equal. But the angle D T Y is equal (I. 15) to the angle GTS. Wherefore in the triangles DTY and GTS there are two angles in the one equal to two angles in the other, and one side equal to one side, opposite to equal angles in each, viz., D Y to GS; for they are the halves of D E and B G. Therefore the remaining sides are equal (I. 26), each to each. Wherefore DT is equal to T G, and Y T to T S. Therefore, if in a parallelopiped, &c. Q. E.D.

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PROP. XL. THEOREM.

If two triangular prisms of the same altitude, have the base of one a parallelo. gram, and the base of the other a triangle; and if the parallelogram be double of the triangle, the prisms are equal.

Let the prisms ABCDEF and GHKLMN be of the same altitude, of which the first is contained by the two triangles ABE and CDF, and the three parallelograms AD, DE, and EC; and the other by the two triangles GHK and LMN, and the three parallelograms LH, HN, and NG; and let one of them have a parallelogram A F, and the other a triangle GHK, for its base: if the parallelogram AF be double of the triangle GHK, the prism ABCDEF is equal to the prism GHKLMN. Complete the solid parallelopipeds A X and GO. Because the parallelogram AF is

double of the triangle GHK; and

B

E

D

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H

F

M

N

G K

the parallelogram HK double (I. 34) of the same triangle. Therefore the parallelogram AF is equal to HK. But parallelopipeds upon equal bases, and of the same altitude, are equal(XI. 31) to one another. Therefore the solid AX is equal to the solid GO. But the prism ABCDEF is half (XI. 28) of the solid AX; and the prism GHKLMN half (XI. 28) 01 the solid GO. Therefore the prism A BCDEF is equal to the prism GHKLMN. Wherefore, if two triangular prisms, &c. Q. E. D.

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