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Then because BC is drawn parallel to the Side

DE, of the Triangle ADE, it fhall be as AB is to of this
BD, fo is AC to CE. But BD is equal to, AC,
Hence as AB is to A C, fo is AC to CE. There
fore a third proportional CE is found to two given
Right Lines AB, AC; which was to be done.

PROPOSITION XII,

PROBLEM.

Three Right Lines being given, to find a fourth proportional to them.

LET A,B,C be three Right Lines given. It is requir'd to find a fourth proportional to them.

Let DE and DF be two Right Lines, making any Angle EDF with each other. Now make DG equal to A, GE equal to B, DH equal to C, and draw the Line GH, as alfo *EF through E, parallel * 31. ↑ to GH.

Then because G H is drawn parallel to EF, the Side of the Triangle DEF, it fhall be as DG is to GE, fo is DH to HF. But DG is equal to A, GE to B, and DH tó C. Confequently as A is to B, fo is C to HF. Therefore the Right Line HF, a fourth Proportional to the three given Right Lines A, B, C, is found; which was to be done.

PROPOSITION XIII.

PROBLEM.

To find a Mean proportional between two given Right
Lines.

LET the two given Right Lines be AB, BC. It
is requir'd to find a Mean proportional between
them. Place AB, BC, in a direct Line, and on the
whole A C defcribe the Semicircle ADC, and draw * * 11. 1.
BD at Right Angles to AC from the Point B, and

let AD, DC, be joined.

Then because the Angle ADC, in a Semicircle, ista Right Angle, and fince the Perpendicular † 31. 31 DB is drawn from the Right Angle to the Bafe; therefore

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*Gor. 8 of therefore DB is a Mean Proportional between the Segments of the Bafe AB, BC. Wherefore a Mean proportional between the two given Lines AB, BC, is found; which was to be done.

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14. 1.

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1 of this.

PROPOSITION XIV.

THEOREM.

Equal Parallelograms having one Angle of the one equal to one Angle of the other, have the Sides about the equal Angles reciprocal; and thofe Parallelograms that have one Angle of the one equal to one Angle of the other, and the Sides that are about the equal Angles reciprocal, are equal between themselves.

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ET AB, BC, be equal Parallelograms, having the Angles at B equal; and let the Sides DB, BE, be in one ftrait Line; then alfo will* the Sides FB, BC, be in one ftrait Line. I fay, the Sides of the Parallelograms AB, BC, that are about the equal Angles, are reciprocal; that is, as DB is to BE, fo is GB to BF.

For let the Parallelogram FE be compleated.

Then because the Parallelogram AB is equal to the Parallelogram BC, and FE is fome other Parallelogram; it fhall be as AB is to FE, fo is BC to FE; but as AB is to FE, fo is DB to BE; and as BC is to FE, fo is GB to BF. Therefore, as DB is to BE, fo is G B to BF. Wherefore the Sides of the Parallelograms AB, BC, that are about the equal Angles, are reciprocally proportional.

And if the Sides that are about the equal Angles are reciprocally proportional, viz. if BĎ be to BE as GB is to B F. I fay the Prallelogram AB is equal to the Parallelogram BC.

For fince DB is to BE as GB is to BF, and DB to BE as the Parallelogram AB to the Parallelogram F E, and GB to BF as the Parallelogram BC to the Parallelogram FE; it fhall be as A B is to FE, fo is BC to FE. Therefore the Parallelogram AB is equal to the Parallelogram BC. And fo equal Parallelograms having one Angle of the one equal to one Angle of the other, have the Sides about the equal An

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gles reciprocal; and thofe Parallelograms that have one Angle of the one equal to one Angle of the other, and the Sides that are about the equal Angles reciprocal, are equal between themselves; which was to be demonftrated.

PROPOSITION XV.

THEOREM.

Equal Triangles having one Angle of the one equal to one Angle of the other, have their Sides about the equal Angles reciprocal; and thofe Triangles that have one Angle of the one equal to one Angle of the other, and have alfo the Sides about the equal Angles reciprocal, are equal between themselves.

LET the equal Triangles ABC, ADE, have one

Angle of the one equal to one Angle of the other, viz the Angle BAC equal to the Angle DAE. I fay the Sides about the equal Angles are reciprocal, that is, as CA is to AD, fo is EA to AB.

For place CA and AD in one ftrait Line, then EA and AB fhall be * also in one ftrait Line, and let * 14. 1. BD be joined. Then because the Triangle ABC is equal to the Triangle A DE, and ABD is fome other Triangle, the Triangle CAB fhall be t to the Tri- † 7.5. angle BAD, as the Triangle ADE is to the Triangle BAD. But as the Triangle CAB is to the Triangle BAD, fo is CA t to AD; and as the Tri- 1 of this: angle EAD is to the Triangle BAD, fo is ‡ EA to AB. Therefore as CA is to AD, fo is E A to AB. Wherefore the Sides of the Triangles ABC, ADE, about the equal Angles, are reciprocal.

And if the Sides about the equal Angles of the Triangles ABC, ADE, be reciprocal, viz. if CA be to AD as EA is to AB. I fay the Triangle ABC is equal to the Triangle ADE.

For, again let BD be joined. Then because CA is to AD as EA is to AB, and CA to AD as the Triangle ABC to the Triangle BAD, and EA to AB as the Triangle EAD to the Triangle BAD; therefore, as the Triangle ABC is to the Triangle BAD, fo fhall the Triangle E AD be to the Triangle

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BAD.

BAD. Whence the Triangles ABC, A DE, have the fame Proportion to the Triangle BAD: And fo the Triangle ABC is equal to the Triangle ADE. Therefore equal Triangles having one Angle of the one equal to one Angle of the other, have their Sides about the equal Angles reciprocal; and thofe Triangles that have one Angle of the one equal to one Angle of the other, and have alfo the Sides about the equal Angles reciprocal, are equal between themfelves; which was to be demonftrated.

PROPOSITION XVI.

THE ORE M.

If four Right Lines be proportional, the Rectangle contained under the Extremes is equal to the Rectangle contained under the Means; and if the Rectangle contained under the Extremes be equal to the Rectangle contained under the Means, then are the four Right Lines proportional.

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ET four Right Lines AB, CD, E, F, be proportional, fo that AB be to CD, as E is to F. I fay the Rectangle contain'd under the Right Lines A B and F, is equal to the Rectangle contain❜d under the Right Lines CD and E.

For draw AG, CH, from the Points A, C, at Right Angles to AB and CD, and make AG equal to F, and CH equal to E, and let the Parallelograms BG, DH, be compleated.

Then because AB is to CD as E is to F, and fince CH is equal to F, and AG to F, it fhall be as AB is to CD, fo is CH to AG. Therefore the Sides that are about the equal Angles of the Parallelograms BG, DHare reciprocal; and fince thofe Parallelograms are *14 of this. equal *, that have the Sides about the equal Angles reciprocal. Therefore the Parallelogram BG is equal to the Parallelogram D H. But the Parallelogram BG is equal to that contain'd under A B and F; for AG is equal to F, and the Parallelogram DH equal to that contain❜d under CD and E, fince CH is equal to E. Therefore the Rectangle contain❜d under AB and F is equal to that contain'd under CD and E.

And

And if the Rectangle contained under A B and F, be equal to the Rectangle contained under CD and E, I fay the four Right Lines are Proportionals, viz. as AB is to CD, fo is E to F.

For, the fame Conftruction remaining, the Rectangle contained under AB and F is equal to that contained under CD and E; but the Rectangle contained under AB and F is the Rectangle BG; för AG is equal to F: And the Rectangle contained under CD and E is the Rectangle DH, for CH is equal to E. Therefore the Parallelogram BG, fhall be equal to the Parallelogram DH, and they are equiangular; but the Sides of equal and equiangular Parallelograms, which are about the equal Angles, are * reciprocal. Wherefore as AB is to CD, fo is CH 14 of this. to AG; but CH is equal to E, and AG to F; there-. fore as AB is to CD, fo is E to F. Wherefore, if four Right Lines be proportional, the Rectangle contained under the Extremes, is equal to the Rectangle contained under the Means; and if the Rectangle contained under the Extremes be equal to the Rectangle contained under the Means, then are the four Right Lines proportional; which was to be demonstrated,

PROPOSITION XVII.

THEOREM.

If three Right Lines be proportional, the Rectangle contained under the Extremes, is equal to the Square of the Mean; and if the Rectangle under the Extremes, be equal to the Square of the Mean, then the three Right Lines are proportional,

LET there be three Right Lines A, B, C, proportional; and let A be to B, as B is to C. I fay, the Rectangle contained under A and C, is equal to the Square of B.

For make D equal to B.

Then because A is to B as B is to C, and B is equal to D, it fhall be as A is to B, fo is D to C. * 7. 5. But if four Right Lines be Proportionals, the Rectan

gle contained under the Extremes is equal to the +16 of this. Rectangle under the Means. Therefore the Rectan

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gle

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