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Confequently the Bafe EF is equal to the Bafe FG, and the Triangle DEF equal to the Triangle GDF; and the other Angles of the one, equal to the other Angles of the other, each to each; under which the equal Sides are fubtended. Therefore the Angle DFG is equal to the Angle DFE, and the Angle G, equal to the Angle E; but the Angle DF G, is equal to the Angle ACB: Wherefore the Angle ACB is equal to the Angle DFE; and the Angle By Hyp. BAC is falfo equal to the Angle EDF: Therefore the other Angle at B is equal to the other Angle at E; and fo the Triangle ABC is equiangular to the Triangle DEF. Therefore, if two Triangles have one Angle of the one, equal to one Angle of the other; and if the Sides about the equal Angles be proportional, ther the Triangles are equiangular; and have thofe Angles equal, under which are fubtended the homologous Sides which was to be demonftrated.

PROPOSITION VII.

THEOREM.

If there are two Triangles, having one Angle of the one equal to one Angle of the other, and the Sides about other Angles proportional; and if the remaining third Angles are either both lefs, or both not less than Right Angles, then shall the Triangles be equiangular and have thofe Angles equal, about which are the proportional Sides.

LET two Triangles ABC, DEF, have one An

gle of the one, equal to one Angle of the other, viz. the Angle B A C equal to the Angle EDF; and let the Sides about the other Angles ABC, DEF, be proportional; viz. as DE is to EF, fo let AB be to BC; and let the other Angles at C and F, be both lefs, or both not lefs than Right Angles. I fay, the Triangle ABC is equiangular to the Triangle DEF; and the Angle ABC is equal to the Angle DEF as alfo the other Angle at C, equal to the other Angle at F.

;

For if the Angle ABC be not equal to the Angle DEF, one of them will be the greater, which let be

ABC.

ABC. Then at the Point B, with the Right Line
AB, make the Angle ABG equal to the Angle* 23. 1.
DEF.

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Now because the Angle A is equal to the Angle D, and the Angle ABG, equal to the Angle DEF; the remaining Angle AGB, ist equal to the remain-t Cor. 32, Ip ing Angle DFE: And therefore the Triangle ABG, is equiangular to the Triangle DEF; and fo as AB is to BG, fo istDE to EF; but as DE is to EF, fo is AB to BC. Therefore as AB is to BC, fo is AB to BG; and fince A B has the fame Proportion to BC, that it has to BG, BC fhall be equal to † 95+ BG; and confequently the Angle at C equal to the Angle BGC. Wherefore each of the Angles BCG, or BGC is lefs than a Right Angle; and confequently, A G B is greater than a Right Angle. But the Angle AGB has been proved equal to the Angle at F; therefore the Angle at F, is greater than a Right Angle: But (by the Hyp.) it is not greater, fince C is not greater than a Right Angle, which is abfurd. Wherefore the Angle ABC is not unequal to the Angle DEF; and fo it must be equal to the fame; but the Angle at A is equal to that at D; wherefore the Angle remaining at C is equal to the remaining Angle at F; and confequently the Triangle ABC is equiangular to the Triangle DEF. Therefore, if there are two Triangles having one Angle of the ene, equal to one Angle of the other, and the Sides about other Angles proportional; and if the remaining third Angles are either both lefs, or both not less than Right Angles, then fhall the Triangles be equiangular; and bave thofe Angles equal, about which are the proportional Sides; which was to be demonftrated.

PRO:

PROPOSITION VIII.

THEOREM.

If a Perpendicular be drawn, in a Right-lined Triangle, from the Right Angle to the Bafe, then the Triangles on each Side of the Perpendicular are fimilar both to the whole, and also to one another...

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ET ABC be a Right-angled Triangle, whose Right Angle is BAC; and let the Perpendicular AD be drawn from the Point A to the Base B C. I fay, the Triangles ABD, ADC, are similar to one another, and to the whole Triangle ABC.

For because the Angle BAC is equal to the Angle ADB, for each of them is a Right Angle, and the Angle at B is common to the two Triangles A BC, Cor. 32. 1. ABD, the remaining Angle ACB fhall be equal to the remaining Angle BAD. Therefore the Triangle ABC is equiangular to the Triangle A BD; and fo † 4 of this, as † BC, which fubtends the Right Angle of the Triangle ABC, is to BA, fubtending the Right Angle of the Triangle ABD, fo is AB fubtending the Angle C of the Triangle ABC to DB, fubtending an Angle equal to the Angle C, viz. the Angle BAD, of the Triangle ABD. And fo moreover is AC to AD, fubtending the Angle B, which is common to the two Triangles. Therefore the Triangle ABC 1 Def. 1 of is equiangular to the Triangle ABD; and the Sides about the equal Angles are proportional. Wherefore the Triangle ABC is ‡ fimilar to the Triangle ABD. I By the fame way we demonftrate, that the Triangle ADC is alfo fimilar to the Triangle ABC. Wherefore each of the Triangles ABD, AD C, is fimilar to the whole Triangle.

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I fay, the faid Triangles are alfo fimilar to one another.

For because the Right Angle BDA is equal to the Right Angle ADC, and the Angle BAD has been proved equal to the Angle C; it follows, that the remaining Angle at B fhall be equal to the remaining Angle DAC. And fo the Triangle ABD is equi

angular

angular to the Triangle ADC. Wherefore as †BD † 4 of this. fubtending the Angle BAD of the Triangle ABD is to DA, fubtending the Angle at C of the Triangle ADC, which is equal to the Angle BAD, fo is AD fubtending the Angle B of the Triangle ABD to DC, fubtending the Angle DAC equal to the Angle B. And moreover, fo is BA to AC, fubtending the Right Angles at D; and confequently the Triangle ABD is fimilar to the Triangle ADC. Wherefore, if a Perpendicular be drawn, in a Right-angled Triangle, from the Right Angle to the Bafe, then the Triangles on each Side of the Perpendicular are fimilar both to the whole, and alfo to one another; which was to be demonftrated.

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Coroll. From hence it is manifeft, that the Perpendicular drawn in a Right-angled Triangle from the Right Angle to the Bafe, is a mean Proportional between the Segments of the Base. Moreover, either of the Sides containing the Right Angle is a mean Proportional between the whole Bafe, and that Segment thereof which is next to the Side.

PROPOSITION IX.

PROBLEM.

To cut off any Part required from a given Right
Line.

ET AB be a Right Line given; from which muft be cut off any required Part; fuppofe a third. Draw any Right Line AC from the Point A, making an Angle at Pleasure with the Line AB. Affume any Part D in the Line AC, make* DE, EC, * 3. 1. each equal to AD, join BC, and draw †DF thro' † 31. 1. D, parallel to BC.

Then because FD is drawn parallel to the Side BC of the Triangle ABC, it fhall be as CD is 2 of this. to DA, fo is BF to FA. But CD is double to DA. Therefore BF fhall be double to FA; and fo BA is triple to AF. Wherefore there is cut off AF, a third Part required of the given Right Line which was to be done.

AB;

PRO

31. I.

34. I

PROPOSITION X.

PROBLEM.

To divide a given undivided Right Lines as an-
other given Right Line is divided.

ET AB be a given undivided Right Line, and
A

AC a divided Line. It is required to divide AB, as AC is divided.

Let A C be divided in the Points D and E, and fo placed, as to contain any Angle with AB. Join the Points C and B; thro' D and E let DF, EG, be drawn * parallel to BC; and thro' D, draw DHK, parallel to A B...

Then FH, HB, are each of them, Parallelograms; and fo DH ist equal to FG, and HK to GB. And because HE is drawn parallel to the Side KC, of the 2 of this. Triangle DKC, it fhall be as CE is to ED; fo is

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KH to HD. But KH is equal to BG, and HD to GF. Therefore, as CE is to ED, fo is BG to GF. Again, because FD is drawn parallel to the Side EG, of the Triangle AGE, as ED is to DA, fo fhall GF be to FA. But it has been proved, thát CE is to ED as BG is to GF. Therefore, as CE is to ED, fo is BG to GF; and as ED is to DA, fo is GF to FA. Wherefore the given undivided Line AB, is divided as the given Line AC is; which was to be done.

PROPÓSÍTION XI.

PROBLEM.

Two Right Lines being given, to find a third proportional to them.

LET AB, AC, be two given Right Lines, fo placed, as to make any Angle with each other." It is required to find a third proportional to AB, AC. Produce AB, AC, to the Points D and E; make BD equal to A C, join the Points B, C, and draw *the Right Line DE thro' D parallel to BC.

Then

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