Then because BC is drawn parallel to the Side DE, of the Triangle ADE, it fhall be as AB is to 2 of this. BD, fo is AC to CE. But BD is equal to AC. Hence as AB is to AC, fo is AC to CE. Therefore 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.

LE
L'equired to find a fourth Lionel them.
ET A, B, C, be three Right Lines given. It is
required 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 thro' E, parallel to GH.

Then because GH 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 to 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.

ET the two given Right Lines be AB, BC. It

them. Place AB, BC, in a direct Line, and on the whole AC defcribe the Semicircle ADC, and* draw * 11. 16 BD at Right Angles to AC from the Point B, and let AD, DC, be joined.

Then because the Angle ADC, in a Semicircle, ist a Right Angle, and fince the Perpendicular † 31 3 DB is drawn from the Right Angle to the Bafe;

therefore

this.

*Cor. 3. of therefore DB is a mean Proportional between the Segments of the Base AB, BC. Wherefore a mean Proportional between the two given Lines AB, BC, is found; which was to be done.

* 14. I.

47.5.
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 those 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.

ET. AB, BC, be equal Parallelograms, having the Angles at B equal; and let the Sides DB, BE, be in one strait Line; then alfo will* the Sides F B, BG, be in one strait 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 some 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 B F. Therefore, as DB is to BE, fo is GB 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 BD be to BE as GB is to BF: I fay the Parallelogram 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 FE, and GB to BF as the Parallelogram BC to the Parallelogram FE; it fhall be as AB 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

gles

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 Angle 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 strait Line, and let * 14. 1. BD be joined. Then because the Triangle ABC is equal to the Triangle ADE, and ABD is fome other Triangle, the Triangle CAB fhall be † to the Tris † 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 to AĎ; 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 EA 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 becaufe 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, so shall the Triangle EAD be to the Triangle M BAD.

BAD. Whence the Triangles ABC, ADE, 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.

THEOREM.

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.

LET four Right Lines AB, CD, E, F, be proporti

onal, fo that AB be to CD, as E is to F. I fay, the Rectangle contained under the Right Lines AB and F, is equal to the Rectangle contained under the Right Lines CD and E.

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

Then because AB is to CD as E is to F, and fince CH is equal to E, 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, DH are 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 DH. But the Parallelogram BG is equal to that contained under AB and F; for AG is equal to F, and the Parallelogram DH equal to that contained under CD and E, fince CH is equal to E. Therefore the Rectangle contained under AB and F is equal to that contained 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; for 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; therefore 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 demonftrated.

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.

ET there be three Right Lines A, B, C, probe portional; 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.

7. 5.

Then because A is to B as B is to C, and B is equal to D, it shall be as A is to B, fo is D to C. But if four Right Lines be Proportionals, the Rectangle contained under the Extremes is † equal to the † 16 of this, Rectangle under the Means. Therefore the RectanM 2.

gle