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equal to B, and DH equal to C; join GH, and through E draw EF parallel to it: FH shall be a

*31. 1. fourth proportional to A,B,C. Because GH is parallel to EF, one of the sides of the triangle DEF, DG: GE :: DH: HF; but DG is

D

A

B

H

equal to A, GE to B, and DH to C; therefore, A : B :: C: HF. Wherefore to the three given straight lines A, B, C, a fourth proportional HF is found. Which was to be done.

PROP. XIII. PROB.

To find a mean proportional between two given straight

lines.

Let the parts AB, BC be equal to the two given straight lines it is required to find a mean proportional between them.

*11. 1.

Upon AC describe the semicircle ADC, and from the point B draw* BD_at_right angles to AC: BD shall be a mean proportional between AB and BC.

Draw AD, DC: and because the angle ADC in a semicircle is a right angle, and because in the

#31. 1.

right angled triangle ADC, BD is drawn from the right angle perpendicular to the base, DB is a mean proportional between AB, BC, the #8. 6. Cor. segments of the base:* therefore between the two given straight lines AB, BC, a mean proportional DB is found. Which was to be done.

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.

Let AB, BC be equal parallelograms, which have the angles at B equal: the sides of the parallelograms AB, BC about the equal angles, shall be reciprocally proportional; that is, DB shall be to BE, as GB to BF.

#14. 1.

Let the sides DB,BE be placed in the same straight line; then FBD, FBE are together equal to two right angles; but by hypothesis FBD is equal to GBE; therefore GBE, FBE are together equal to two right angles; wherefore also FB, BG are in one straight line.* Complete the parallelogram FE: and because the parallelogram AB is equal to BC, and that FE is another parallelogram, AB: FE:: BC: FE; but AB: FE: DB: BE, and as BC FE :: GB: BF;

1. 6.

#2. 5.

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therefore DB: BE:: GB:

BF. Wherefore, the sides of the parallelograms AB, BC about their equal angles are reciprocally proportional.

Next, let the sides about the equal angles be reciprocally proportional, viz. as DB to BE, so GB to BF: the parallelogram AB shall be equal to the parallelogram BC. Because, DB: BE:: GB: BF; and DB: BE:: AB:

*2. 6.

*2.5. #9.5.

Cor. 1.

FE; and GB: BF:: BC: FE; therefore AB: FE: BC: FE: therefore the parallelogram AB is equal to the parallelogram BC. Therefore equal parallelograms, &c. Q. E. D.

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 shall be reciprocally proportional; that is, CA shall be to AD, as EA to AB.

Let the triangles be placed so that their sides CA, AD may be in one straight line; then it may be proved as in last proposition, that EA and AB

are in one straight line;* join #14. 1. B, D. Because the triangle ABC is equal to the triangle ADE, and

E

that ABD is another triangle: therefore triangle CAB: triangle BAD :: triangle AED : triangle DAB; but triangle CAB: triangle BAD :: CA: AD,* and triangle EAD: triangle DAB :: EA: AB;* therefore CA: AD: EA: AB:* wherefore the sides of the triangles ABC, ADE, about the equal angles are reciprocally proportional.

#1. 6. #1.6. *9. 5.

Cor. 1.

Next, 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 shall be equal to the triangle ADE.

#1. 6.

#1. 6.

*2. 5.

Join BD as before: then because, CA: AD :: EA: AB; and CA: AD :: triangle ABC: triangle BAD; and EA AB: triangle EAD: triangle BAD;* therefore* triangle BAC : triangle BAD :: triangle EAD triangle BAD; and the consequents being equal, the antecedents are equal:* wherefore the triangle ABC is equal to the triangle ADE. Therefore equal triangles, &c. Q. E. D.

#9.5. Cor. 1

PROP. XVI. THEOR.

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.

*11. 1.

+3. 1.

+31. 1.

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 shall be equal to the rectangle contained by CD, E. From the points A, C draw* AG, CH at right angles to AB, CD; and maket AG equal to F, and CH equal to E: and complete the parallelograms BG, DH. Because, AB: CD :: E: F; and that E is equal to CH, and F to AG; AB : CD :: CH : 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 are equal to one reciprocally proportional, another; therefore the parallelogram BG is equal to the parallelogram DH: but 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.

*14. 6.

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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 shall be proportional, viz. AB shall be 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 rect

+1 Ax.

angle 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:* wherefore, AB : CD :: CH: AG: but CH is equal to E, and AG to F; therefore AB CD :: E: F. Wherefore if four, &c.

*14. 6.

2. E. D.

NOTE. The proposition next following is an obvious corollary from this, being that particular case of it in which the means are equal. The 17th may therefore be omitted as superfluous, but the corollary which we have subjoined to it is of importance in the subsequent demonstrations.

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 shall be equal to the square of B.

Take D equal to B: and because A: B:: B: C, and that B is equal D; A: B:: D: C: but if four straight lines be proportionals, the rect

*16. 6.

angle contained by the extremes is equal to that which is contained by the means;* 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

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