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; which was to be demonftrated.

PROPOSITION XV.
THEOREM.

Equal Triangles, baving 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 also
the Sides about the equal Angles reciprocal, are
equal.

L

ET the equal Triangles A B C, ADE, have one Angle of the one equal to one Angle of the other, viz. the Angle B A C 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 A B fhall be alfo in one ftrait Line; and let BD be joined. Then, because the Triangle A B C is equal to the Triangle A DE, and A B D is fome other Triangle, the Triangle C A B fhall be + to the Triangle BA D, as the Triangle ADE is to the Triangle B AD. But, as the Triangle CAB is to the Triangle BAD, fo is CA to AD; and as the Triangle E A D is to the Triangle BAD fo is EA to AB. Therefore, as C A is to A D,* fo is EA to A B. Wherefore the Sides of the Triangles A B C, ADE, about the equal Angles are reciprocal.

And, if the Sides about the equal Angles of the Triangles A B C, ADE, be reciprocal, viz. if CA be to A D, as E A is to AB; I fay, the Triangle A BC is equal to the Triangle ADE.

For, again, let BD be joined. Then, becaufe CA is to A D, as EA is to AB; and CA to AD ‡, as the Triangle ABC to the Triangle B AD; and EA to AB, as the Triangle E AD to the Triangle BAD; therefore, as the Triangle ABC is to the

Triangle

11. 5.

Triangle B A D,* fo fhall the Triangle E A D be to
the Triangle B AD. Whence the Triangles A B C,
ADE, have the fame Proportion to the Triangle
BAD; and fo the Triangle A B C is equal to the ‡ 7.5.
Triangle A D E. 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 An-
gle of the other, and have alfo the Sides about the equal
Angles reciprocal, are equal; which was to be demon-
ftrated.

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 A B, C D, E, F, be pro

portional, fo that A B be to CD, as E is to F. I fay, the Rectangle contained under the Right Line AB and F, is equal to the Rectangle contained under the Right Lines C D and E.

For, draw A G and C H, from the Points A and C, at Right Angles to A B and CD; and make AG equal to F, and C H equal to E; and let the Parallelograms BG, D H be compleated.

Then, becaufe A B is to CD, as E is to F; and fince C H is equal to E, and AG to F; it fhall be, as A B is to CD, fo is CH to AG. Therefore, the Sides that are about the equal Angles of the Parallelograms B G, D H, are reciprocal; and fince those Parallelograms are equal *, that have the Sides about 14 of this. the equal Angles reciprocal; therefore the Parallelogram B G is equal to the Parallelogram DH. But the Parallelogram B G is equal to that contained under AB and F; for A G is equal to F, and the Parallelogram D H equal to that contained under CD and E, fince CH is equal to E, Therefore the Rectangle M 3 contained

contained under A B and F, is equal to that contained under C D and E.

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 A B is to CD, fo is E to F.

For the fame Conftruction remaining, the Rectangle contained under A B and F is equal to that contained under CD and E; but the Rectangle contained under A B and F is the Rectangle B G; for A G 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 B G 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 14 of this.reciprocal. Wherefore, as A B is to CD, fo is AB CH to AG; but C H is equal to E, and A G to F; therefore, as A B 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.

#7.5.

PROPOSITION XVII.
THEORE M.

If three Right Lines be proportional, the Reɛtan-
gle 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.

L

E T 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 Bfo is D to C. But, if four Right Lines be Proportionals, the Rectan

gle

gle contained under the Extremes is + equal to the t 16 of this. Rectangle under the Means. Therefore, the Rectangle contained under A and C is equal to the Rectangle under B and D: But the Rectangle under B and D is equal to the Square of D; for B is equal to D: Wherefore the Rectangle contained under A, C, is equal to the Square of B.

And if the Rectangle contained under A and C be equal to the Square of B: I fay, as A is to B, fo is B to C.

For the fame Conftruction remaining, the Rectangle contained under A and C, is equal to the Square of B; but the Square of B is the Rectangle contained under B and D; for B is equal to D; and the Rectangle contained under A and C fhall be equal to the Rectangle contained under B and D. But if the Rectangle contained under the Extremes, be equal to the Rectangle contained under the Means, the four Right Lines fhall be Proportionals. Therefore A is 16 of this. to B as D is to C; but B is equal to D. Wherefore A is to B, as B is to C. Therefore, 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; which was to be demonftrated.

*

PROPOSITION XVIH.

PROBLEM.

Upon a given Right Line, to defcribe a Rightlined Figure, fimilar, and fimilarly fituate, to a Right-lined Figure given.

LET AB be the Right Line given, and CE the Right-lined Figure. It is required to defcribe upon the Right Line A B a Figure fimilar, and fimilarly fituate, to the Right-lined Figure CE.

Join DF, and make, * at the Points A and B, with 3. x. the Line A B, the Angles GA B, ABG, feverally equal to the Angles C and CDF. Whence the other Angle CFD is equal to the other Anglet Cor. AG B; and fo the Triangle F C D is equiangular to the Triangle GAB: And, confequently, as FD is

32.1.

$4 of this

11. 5.

to G B, fo is 1 FC to GA; and fo is CD to A B. Again, make the Angles B GH, GBH, at the Points Band G, with the Right Line B G, feverally equal to the Angles EFD, EDF; then the remaining An+Cor. 32.1. gle at E is equal to the remaining Angle at H. Therefore the Triangle F D E is equiangular to the Triangle G BD; and confequently, as F D is to H4 of this. GB, fo is FE to GH; and fo ED to H B. But 11 it has been proved, that F D is to G B, as FC is to GA, and as CD to AB. And therefore, as FC is to AG, fo is CD to A B; and fo F E to GH; and fo ED to HB. And because the Angle C F D is equal to the Angle AGB'; and the Angle DFE equal to the Angle BGH: the whole Angle CFE fhall be equal to the whole Angle AG H. By the fame Reason, the Angle C D E is equal to the Angle ABH; and the Angle at C equal to the Angle at A; and the Angle E equal to the Angle H. Therefore the Figure A His equiangular to the Figure CE; and they have the Sides about the equal Angles pro+Def. of portional. Confequently, the Right-lined Figure A H will be fimilar to the Right-lined Figure CE. Therefore, there is defcribed upon the given Right Line A B the Right lined Figure A H, fimilar, and fimilarly fituate, to the given Right lined Figure CE; which was to be done.

this.

PROPOSITION XIX.
THEOREM.

Similar Triangles are in the duplicate Proportion
of their bomologous Sides.

L

ET ABC, DEF, be fimilar Triangles, having the Angle B equal to the Angle E; and let A B be to B C, as D E is to EF, so that B C be the Side homologous to EF. I fay, the Triangle A B C, to the Triangle DEF, has a duplicate Proportion to that of the Side B C to the Side E F.

Ty ofthis. For, take * BG a third Proportion to B C and EF; that is, let BC be to EF as EF is to BG, and join GA.

Then, because A B is to B C, as DE is to EF; it fhall be (by Alternation), as A B is to DE, fo is BC to E F; but as B C is to E F, fo is EF to B G. Therefore, as A B is to DE, fo ist EF to BG: Confequently,