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are therefore four magnitudes, viz. the two bases AF, FH, and the two folids A Y, Y H. and there are taken equimul. tiples of the base A F, and the folid A Y, víz. the bafe LF, and the folid LY, and the base NF, and the folid NY: And it has been demonftrated, if the bafe L F exceeds the bafe N F, the folid L Y does exceed the folid NY; and if it be equal, equal; if lefs, lefs. Therefore [by 5. def. 5.1 as the base AF is to the base F H, fo is the folid AY to the folid Y H. Which was to be demonstrated.

The demonftration of this theorem is another proof of the existence and truth of the fifth definition of the fifth book.

PROP. XXVI.

PRO B L.

To make a folid angle at a given right line, and at a given point in it, equal to a given folid angle. ・

Let the given right line be a B, and A the given point in it: and let the given folid angle be at D, contained under the three plane angles E DC, EDF, FDC: it is required to make a folid angle at the given right line A B : and at the given point a in it, equal to the folid angle

at. D.

Take any point F in the line D F, from which [by 11. 11.] draw F G perpendicular to the plane paffing thro' ED, DC, meeting the plane in the point G, and join DG, Again make [by 23. 1.] the angle B A L, at the given right line AB, and at the given point A in it, equal to the angle EDC, and the

angle B'A K equal to the angle EDG: and [by 3. 1.] make AK equal to DG, and [by 12. 11.] draw KH from K at right angles to the plane paf

fing thro' BAL, and

B

A

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F

make KH equal to GF, and join HA: I fay the folid añgle at A, which is contained under the angles BAL, BHA, HAL is equal to the folid angle at D comprehended under the angles EDC, EDF, FDC.

Book XI. For take the equal right lines A B, DE; and join H B, KB, FE, GE. Then because F G is perpendicular to the plane drawn thro' ED, DC; it will be perpendicular to all right lines that touch it, and are in that plane [by 3. def. 11:] Therefore the angles FGD, FGE are each of them a right angle. By the fame reafon the angles HKA, HKB are each of them a right angle. And because the two fides KA, AB are equal to the two fides GD, DE, each to each, and they contain equal angles; the bafe BK will be [by 4. 1.] equal to the bafe E G. But K H is equal to GF, and they contain right angles: Therefore HB is equal to FE. Again, because the two fides A K, K H are equal to the two fides DG, GF, and they contain equal angles, the base AH will be equal to DF. But A B is equal to DE: Therefore the two fides HA, A B are equal to the two fides F D, DE, and the base H B is equal to the base FE: Therefore the angle BAH will be equal to the angle EDF: By the fame reason the angle HAL is equal to the angle FDC, viz. if we take A L, DC equal, and join K L, HL, GC, FC, because the whole angle B A L is equal to the whole angle EDC, and B A K is put equal to EDG; the remaining angle K AL will be equal to the remaining angle G D C, And because the two fides KA, A L, are equal to the two fides GD, DC, and they contain equal angles; the bafe KL [by 4. 1.] will be equal to the base GC. But KH is equal to GF: Therefore the two fides LK, KH are equal to the two fides CG, GF, and they contain right angles; wherefore the bafe HL is equal to the base FC. Again, because the two fides HA, AL are equal to the two fides F D, DC, and the base HL is equal to the bafe FC; the angle HAL [by 8. 1.] will be equal to the angle F D C, and the angle B AL equal to the angle EDC.

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Therefore a folid angle is made at a given right line, and at a given point in it, equal to a given folid angle. Which was to be done.

The folution of the following problem would be new, viz, a folid angle confifting of three equal plane angles being given: To find a folid angle confifting of four equal plane anles, that fhall be equal to it, when poffible.

PROP

PROP. XXII. PRO BL.

Upon a given right line to defcribe a folid parallelepipedon fimilar and alike fituate to a given folid parallelepipedons.

Let the given right line be A B, and the given folid parallelepipedon be DC: It is required to describe a parallelepipedon upon the given right line AB, fimilar and alike fituate to the given folid parallelepipedon DC.

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M

D

For at the given right line A B, and at the given point A in it, make the folid angle which is contained under the plane angles BAH, HAK, KA B, equal to the folid angle c, fo that the angle BAH, be equal to the angle ECF, the angle BAK equal to ECG, and the angle KAH equal to the angle GCF. Then [by 12.6.] make as EC is to CG, fo is BA to A K, and as G C is to CF, fo is KA to A H. Then by equality of ratio [by 22. 5.] as EC is to CF, fo will B A be to AH. Laftly, complete the parallelogram B H and the folid A L.

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Then because it is as EC is to CG, fo is BA to AK, the fides about the equal angles ECG, BA K are proportional and fo [by 4. 6.] the parallelogram GE is fimilar to the parallelogram K B. Allo by the fame reason, the parallelogram K H is fimilar to the parallelogram GF, and the parallelogram F E to the parallelogram HB: Therefore three parallelograms of the folid CD, are fimilar to three parallelograms of the folid A L. But [by 24. 11.] the three oppofite parallelograms are equal and fimilar to them. Therefore the whole folid CD will be fimilar to the whole folid A L.

Therefore upon a given right line A B a folid parallelepidedon A L is defcribed fimilar to a given folid parallelepipendon C D, and alike fituate, Which was to be done.

• Two folids are faid to be alike fituate, when their homologous or answerable planes have like fituations.

PROP.

PROP. XXVIII. THEOR. If a folid parallelepipedon be cut by a plane paffing thro' the diagonals of the oppofite parallelograms; the folid will be cut by that plane into two equal parts.

For let the folid parallelepipedon A B be cut by the plane CDEF paffing thro' the diagonals CF, DE of two oppofite planes: I fay the folid AB is cut in half by the plane

CDEF.

For because [by 34. 1.] the trianBgle cor is equal to the triangle CBF, and the triangle ADE to the triangle DEH, and [by 24. 11.] the parallelogram CA is equal to the parallelogram BE, for it is oppofite: and the Hparallelogram GE is equal to the parallelogram CH: [by 10. def. 11.] the prifm contained under the two triangles C GF, ADE, and the three parallelograms GE, A C, CE is equal to the prifm contained under the two triangles CFB, DE H, and the three parallelograms CH, BE, CE: for they are eontained under equal numbers of equal planes.

A

E

Therefore the whole folid A B is cut in half by the plane CDEF. Which was to be demonftrated.

PROP. XXIX. THEOR.

Solid parallelepipedons, ftanding upon the fame base, and having the fame altitude, and whofe fides Standing upon the common bafe, terminate in the fame right lines, are equal to one another.

For let the folid parallelepipedons CM, CN, of the fame altitude, ftand upon the fame base A B, and let their fides AF, AG, LM, L N, CD, CE, BH, BK terminate in the right lines FN, DK: I fay the folid c M is equal to the folid c N.

For because C H, C K are both parallelograms: [by 34. 1.] CB will be equal to DH, or EK; therefore DH is equal to E K. Take away EH, which is common ;

then

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the remainder D E is equal
to the remainder H K ;
and fo [by 8. 1.] the tri-
angle DEC is equal to the
triangle H K B. But [by
36. 1.] the parallelogram C
DG is equal to the paral-
lelogram H N. By the
fame reason the triangle

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AFG is equal to the triangle L M N. But [by 24. 11.] the parallelogram C is equal to the parallelogram B M, and the parallelogram CG to the parallelogram B N; for they are oppofite. Therefore [by 10. def. 11] the prism contained under the two triangles A F G, D E C, and the three parallelograms A D, G D, G C, is equal to the prifm contained under the two triangles, L MN, HBK, and the three parallelograms BM, NH, B N,. Add the common folid, whose base is the parallelogram A B, and its oppofite is GEHM: Therefore the whole folid parallelepipedon C M is equal to the whole folid parallelepipedon c N.

Therefore folid. parallelepipedons upon the fame base, and having the fame altitude, and whofe fides ftanding up-. on the common bafe, terminate in the fame right lines, are equal to one another, Which was to be demonftrated.

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Solid parallelepipedons, having the fame base and al titude, and whofe fides ftanding upon the common bafe, do not terminate in the fame right lines, are equal to one another.

For let the folid parallelepipedons CM. CN be upon the fame base A B, and have the fame altitude, and let their fides AF, A G, L M, LN, CD, CE, BH, BK, ftanding upon the bafe, not terminate in the fame right lines: I fay the folid CM is equal to the folid c N.

For produce NK, DH; as alfo GE, FM; and let them meet in the points o, R,P, X; and join AX, LO,.

CP, BR.

The folid c M, whose base is the parallelogram A C BL, and oppofite parallelogram F D H M, is [by 29. 11] equal to the folid co, whofe bafe is the parallelogram ACBL,

and

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