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Book VL

PROP. XIV. THEOR.

EQUAL parallelograms which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: and parallelograms that have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another.

F

Let AB, BC be equal parallelograms, which have the angles at B equal, and let the sides DB, BE be placed in the same a 14. 1. straight line; wherefore also FB, BG are in one straight line 2 : the sides of the parallelograms AB, BC about the equal angles, are reciprocally proportional; that is, DB is to BE, as GB to BF. Complete the parallelogram FE; and because the parallelogram AB is equal to BC, and that A FE is another parallelogram, AB is to FE, as BC to FE: but as AB to FE, so is the base DB to BE; and, as BC to FE, so is the base GB to BF; therefore, as DB d 11. 5. to BE, so is GB to BF d. Wherefore the sides of the parallelograms AB, BC about their equal angles are reciprocally proportional.

b 7.5.

c 1. 6.

€ 9. 5.

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But, let the sides about the equal angles be reciprocally proportional, viz. as DB to BE, so GB to BF; the parallelogram AB is equal to the parallelogram BC.

Because, as DB to BE, so is GB to BF; and as DB to BE, so is the parallelogram AB to the parallelogram FE; and as GB to BF, so is the parallelogram BC to the parallelogram FE; therefore as AB to FE, so BC to FEd; wherefore the parallelogram AB is equal to the parallelogram BC. Therefore, equal parallelograms, &c. Q. E. D.

Book VI,

PROP. XV. THEOR.

EQUAL triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: and triangles which have one angle in the one equal to one angle in the other, and their sides about the equal angles reciprocally proportional, are equal to one another.

Let ABC, ADE be equal triangles, which have the angle BAC equal to the angle DAE; the sides about the equal angles of the triangles are reciprocally proportional; that is, CA is to AD, as EA to AB.

Let the triangles be placed so that their sides CA, AD be in one straight line; wherefore also EA and AB are in one straight line; and join BD. Because the triangle ABC is equal to the a 14. 1. triangle ADE, and that ABD is

another triangle; therefore as the triangle CAB is to the triangle BAD, so is triangle EAD to triangle DAB: but as triangle CAB to triangle BAD, so is the base CA to AD ©; and as triangle EAD to triangle DAB, so is the base EA to AB; as therefore CA to AD, so is EA to ABd:

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wherefore the sides of the triangles ABC, ADE about the equal angles are reciprocally proportional,

But let the sides of the triangles ABC, ADE about the equal angles be reciprocally proportional, viz. CA to AD, as EA to ÀB; the triangle ABC is equal to the triangle ADE.

Having joined BD as before; because as CA to AD, so is EĄ to AB; and as CA to AD, so is triangle BAC to triangle BADc; and as EA to AB, so is triangle EAD to triangle BAD; therefored as triangle BAC to triangle BAD, so is triangle EAD to triangle BAD; that is, the triangles BAC, EAD have the same ratio to the triangle BAD: wherefore the triangle ABC is equal e 9. 5 to the triangle ADE. Therefore, equal triangles, &c. Q. E. D.

e

Book VI.

PROP. XVI. THEOR.

a 11. 1.

b 7. 5.

IF four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means: and if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportionals.

Let the four straight lines AB, CD, E, F be proportionals, viz. as AB to CD, so E to F; the rectangle contained by AB, F is equal to the rectangle contained by CD, E.

From the points A, C draw a AG, CH at right angles to AB, CD; and make AG equal to F, and CH equal to E, and complete the parallelograms BG, DH: because as AB to CD, so is E to F; and that E is equal to CH, and F to AG; AB is to CD, as CH to AG: therefore the sides of the parallelograms BG, DH about the equal angles are reciprocally proportional; but parallelograms which have their sides about equal angles c 14. 6. reciprocally proportional, are equal to one another; therefore the parallelogram BG is equal to the parallelogram DH: and the parallelogram BG is contained by the straight lines AB, F, because AG is equal to F; and the parallelogram DH is contained by CD and E, because CH is equal to E: therefore the rectangle contained by the straight lines AB, F is equal to that which is contained by CD and E.

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And if the rectangle contained by the straight lines AB, F be equal to that which is contained by CD, E; these four lines are proportionals, viz. AB is to CD, as E to F.

The same construction being made, because the rectangle contained by the straight lines AB, F is equal to that which is contained by CD, E, and that the rectangle BG is contained by AB, F, because AG is equal to F; and the rectangle DH by CD, E, because CH is equal to E; therefore the parallelogram BG is equal to the parallelogram DH; and they are equiangu

lar: but the sides about the equal angles of equal parallelograms Book VI. are reciprocally proportionale: wherefore, as AB to CD, so is CH

to AG; and CH is equal to E, and AG to F: as therefore AB is c 14. 6. to CD, so E to F. Wherefore, if four, &c. Q. E. D.

PROP. XVII. THEOR.

IF three straight lines be proportionals, the rectangle contained by the extremes is equal to the square of the mean: and if the rectangle contained by the extremes be equal to the square of the mean, the three straight lines are proportionals.

Let the three straight lines A, B, C be proportionals, viz. as A to B, so B to C; the rectangle contained by A, C is equal to the square of B.

Take D equal to B; and because as A to B, so B to C, and that B is equal to D; A is to B, as D to C: but if four straight lines a 7. 51⁄2be proportionals, the rect

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C is equal to that contained by B, D. But the

rectangle contained by B, D is the square of B; because B is equal to D:

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therefore the rectangle contained by A, C is equal to the square of B.

And if the rectangle contained by A, C be equal to the square of B; A is to B, as B to C.

The same construction being made, because the rectangle contained by A, C is equal to the square of B, and the square of B is equal to the rectangle contained by B, D, because B is equal to D; therefore the rectangle contained by A, C is equal to that contained by B, D: but if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportionals: therefore A is to B, as D to

Book VI. C; but B is equal to D; wherefore as A to B, so B to C. Therefore, if three straight lines, &c. Q. E. D.

See N.

a 23. 1.

b 32. 1.

PROP. XVIII. PROB.

UPON a given straight line to describe a rectilineal figure similar, and similarly situated to a given rectilineal figure,

Let AB be the given straight line, and CDEF the given rectilineal figure of four sides; it is required upon the given straight line AB to describe a rectilineal figure similar, and similarly situated to CDEF.

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Join DF, and at the points A, B, in the straight line AB, make the angle BAG equal to the angle at C, and the angle ABG equal to the angle CDF; therefore the remaining angle CFD is equal to the remaining angle AGBb: wherefore the triangle FCD is equiangular to the triangle GAB: again, at the points. G, B, in the straight line GB, make a the angle BGH equal to the angle DFE, and the angle GBH equal to FDE; there

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fore the remaining angle FED is equal to the remaining angle GHB, and the triangle FDE equiangular to the triangle GBH: then, because the angle AGB is equal to the angle CFD, and BGH to DFE, the whole angle AGH is equal to the whole CFE: for the same reason, the angle ABH is equal to the angle CDE; also the angle at A is equal to the angle at C, and the angle GHB to FED: therefore the rectilineal figure ABHG is equiangular to CDEF: but likewise these figures have their sides about the equal angles proportionals: because the triangles GAB, FCD being equiangular, BA is to AG, as DC to CF; and because AG is to GB, as CF to FD; and as GB to GH, so, by reason of the equiangular triangles BGH, DFE, is FD to d 22.5. FE; therefore, ex æqualia, AG is to GH, as CF to FE: in the same manner it may be proved that AB is to BH, as CD to DE: and GH is to HB, as FE to ED. Wherefore, because

c 4. 6.

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