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PROPOSITION X.-PROBLEM.

To divide a given straight line similarly to a given divided straight line, that is, into parts that shall have the same ratios to one another which the parts of the divided given straight line have.

LET ab be the straight line given to be divided, and a c the divided line it is required to divide a b similarly to a c.

Let a c be divided in the points d, e; and let a b, a c be placed so as to contain any angle, and join bc, and through the points d, e, draw (i. 31) df, eg parallels to it; and through d draw dhk parallel to ab: therefore each of the figures fh, hb, is a parallelogram : wherefore dh is equal (i. 34) to fg, and hk to gb and because he is parallel to kc, one of the sides of the triangle dk c, as ce to ed, so is (vi. 2) kh to hd: but kh is equal to bg, and hd to f gf; therefore, as ce to ed, so is bg to gf. Again, because fd is parallel to eg, one of the sides of the triangle age, as ed to da, so is gf to fa: but it has been proved that ce is to ed b as bg to gf; and as ed to da, so is gf to fa: therefore the given straight line ab is divided similarly to a c was to be done.

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PROPOSITION XI.-PROBLEM.

To find a third proportional to two given straight lines.

LET a b, ac be the two given straight lines, and let them be placed so as to contain any angle; it is required to find a third proportional to a b, a c.

Produce a b, a c, to the points d, e; and make bd equal to a c; and having joined bc, through d draw de parallel to it (i. 31).

Because bc is parallel to de, a side of the triangle ade, ab is (vi. 2) to bd, as ac to ce: but bd is equal to ac; as therefore ab to a c, so is a c to ce Wherefore, to the two given straight lines ab, ac, a third proportional ce is found. Which was to be done.

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PROPOSITION XII.-PROBLEM.

To find a fourth proportional to three given straight lines.

LET a, b, c be the three given straight lines; it is required to find a fourth proportional to a, b, c.

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Take two straight lines de, df, containing any angle edf; and upon

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these make dg equal to a, ge equal to b, and dh equal to c; and having joined gh, draw ef parallel (i. 31) to it through the point e and because gh is parallel to ef, one of the sides of the triangle def dg is to ge as dhto hf (vi. 2) ; but dg is equal to a, ge to b, and dh to c; therefore, as a is to b, so is c to h f. Wherefore to the three given straight lines a, b, c, a fourth proportional hf is found. Which was to be done.

PROPOSITION XIII.-PROBLEM.

To find a mean proportional between two given straight lines.

LET a b, bc be the two given straight lines: it is required to find a mean proportional between them.

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Place a b, bc in a straight line, and upon ac describe the semicircle

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adc, and from the point b draw (i. 11) bd at right angles to a c, and join a d, dc.

Because the angle a dc in a simicircle is a right angle (iii. 31), and because in the right-angled triangle a d c, d b is drawn from the right angle perpendicular to the base, db is a mean proportional between ab, bc the segments of the base (vi. 3. cor.): therefore between the two given straight Which was to done.

lines ab, bc, a mean proportional db is found.

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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.

LET ab, bc be equal parallelograms, which have the angles at b equal, and let the sides db, be be placed in the same straight line; wherefore also fb, bg are in one straight line (i. 14). The sides of the parallelo

grams ab, bc about the equal angles, are reciprocally proportional; that

is, db is to be as gb to bf.

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Complete the parallelogram fe; and because the parallelogram ab is equal to bc, and that fe is another parallelogram, a b is to fe as bc to fe (v. 7): a but as ab to fe, so is the base db to be (vi. 1) and as bc to fe, so is the base of gb to bf; therefore, as db to be, so is gb to bf (v. 11). Wherefore, the sides of the parallelograms ab, bc about their equal angles are reciprocally proportional.

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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 a b to fe, so is bc to fe (v. 11): wherefore the parallelogram a b is equal (v. 9) to the parallelogram bc. Therefore equal parallelograms, &c. Q. E. D.

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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, a de 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 a d, as ea to a b.

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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 (i. 14); and join bd. Because the triangle abc is equal to the triangle ade, and that abd is another triangle; therefore as the triangle cab, is to the triangle bad, so is the triangle ead to triangle dab (v. 7): but as triangle cab to triangle bad, so is the base ca to ad (vi. 1); and as triangle ead to triangle da b, so is the base e a to a b (vi. 1): as therefore ca to a d, so is e a to a b (v. 11); wherefore the sides of the triangles a b c, ade about the equal angles are reciprocally proportional.

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But let the sides of the triangles abc, ade about the equal angles be reciprocally proportional, viz. ca to ad, as ea to ab; the triangle abc is equal to the triangle ade.

Having joined bd as before; because, as ca to ad, so is ea to ab; and as ca to a d, so is triangle abc to triangle bad (vi. 1); and as e a to a b, so is triangle e ad to triangle bad (vi. 1); therefore (v. 11) as

triangle bac to triangle bad, so is triangle e ad to triangle bad: that is, the triangles bac, ead have the same ratio to the triangle bad: wherefore the triangle abc is equal (v. 9) to the triangle a de. Therefore equal triangles, &c. Q. E. D.

PROPOSITION XVI.-THEOREM.

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, c d, e, f, be proportionals, viz. as ab to cd, so e to f; the rectangle contained by a b, f is equal to the rectangle contained by cd, e.

From the points a, c, draw (i. 11) ag, ch at right angles to a b, cd; and make ag equal to f, and ch equal to e, and complete the parellelograms bg, dh; because, as a b to cd, so is e to f; and that e is equal to ch, and f to ag; ab is (v. 7) to cd as ch to a g. Therefore the sides of the parallelograms bg, dh about the equal angles are reciprocally proportional; but parallelograms which have their sides about equal angles

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reciprocally proportional, are equal to one another (vi. 14); 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 d straight lines ab, f is equal to that which is contained by cd and e.

b c 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 a b, f, because a g 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 equiangular: but the sides about the equal angles of equal parallelograms are reciprocally proportional (vi. 14): wherefore, as ab to cd, so is ch to ag; and ch is equal to e, and a g to f: as therefore a b is to cd, so is e to f. Wherefore, if four, &c. Q. E. D.

PROPOSITION XVII.—THEOREM.

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.

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Take d equal to b; and because as a to b so b to c, and that bis equal to d: a is (v. 7) to b as d to c: but if four straight lines be proportionals, the rectangle contained by the extremes is equal to that which is contained by the means (vi. 16): therefore the rectangle contained by a, 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 therefore the rectangle contained by a, c is equal to the square of b.

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And if the rectangle contained by a, c be equal to the square of b; a is to b as b is 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 (vi. 16): therefore a is to bas d to 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.

PROPOSITION XVIII.-THEOREM.

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

LET a b 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 c def

Join df, and at the points a, b in the straight line ab make (i. 23) 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 agb (i. 32). Wherefore the triangle fcd is equiangular to the triangle gab: again, at the points g, b in the straight line gb, make (i. 23) the angle bg h equal to the angle dfe, and the angle gbh equal to fde; therefore the remaining angle fed is equal to the remaining angle g hb, and the triangle fde equiangular to the triangle gbh: then, because the

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