angle DGF; and the remaining angles will be equal to the re- Book VI, maining angles, each to each, under which the equal fides are extended; therefore the angle DFG is equal to the angle DFE; and the angle at G to the angle at E; but the angle DFG is equal (by conft.) to the angle ACB; therefore also ACB is equal to DFE: but BAC is also fuppofed equal to EDF; therefore the remaining angle at B is equal to the remaining angle at E: wherefore the triangle ABC is equiangular to the triangle Def.

Wherefore if two triangles have one angle equal to one angle, and the fides about the equal angles proportionals: the triangles will be equiangular, and will have the angles equal, under which the fides of like ratio are extended. Which was to be demonstrated.

If two triangles have one angle equal to one angle; and the fides about two other angles proportionals; and if each of the remaining angles be at the fame time lefs or not less than a right angle: the triangles will be equiangular; and will have thofe angles equal, about which the fides are, which are proportionals.

Let ABC, DEF be two triangles having one angle equal to one angle viz. the angle BAC equal to the angle EDF; and the fides about two other angles proportionals, the angles ABC, DEF; so that AB may be to BC as DE to EF; and first let each of the remaining angles at C and F be at the fame time lefs than a right angle: I fay, that the triangle ABC is equiangular to the triangle DEF; and that the angle ABC will be equal to the angle DEF; and the remaining angle viz. the angle at C equal to the remaining angle at F. For if the angle ABC be unequal to the angle DEF, one of them is greater: let ABC be the greater: and with the straight line AB, and at

the point B in it, let the angle

B

A

D

G

C E

ABG be made equal to the angle DEF.

And because the angle A is equal to D; and the angle ABG to DEF; therefore the remaining angle AGB is equal to the remaining * E

VOL. I.

angle

Book VI. angle DFE: therefore the triangle ABG is equiangular to the triangle DEF; therefore it is (by 4. 6.) as AB is to BG so is DE to EF; but as DE is to EF fo is AB fuppofed to be to BC; and therefore (by 11. 5.) as AB is to BC fo is AB to BG: wherefore AB has the fame ratio to each of the lines BC, BG; therefore (by 9. 5.) BC is equal to BG; fo that alfo (by 5. 1.) the angle BGC is equal to the angle BCG: but the angle at C is supposed to be less than a right angle; therefore BGC is also less than a right angle ; fo that the angle adjacent to it viz. AGB is (by 13. 1.) greater than a right angle; and it has been demonftrated to be equal to the angle at F; therefore the angle at F is greater than a right angle; but it is supposed to be less than a right angle; which is abfurd: wherefore the angle ABC is not unequal to the angle DEF, therefore it is equal and also (by supp.) the angle at A is equal to the angle at D; therefore (by 32. 1.) the remaining angle at C is equal to the remaining angle at F: Therefore the triangle ABC is equiangular to the triangle DEF.

But again, let each of the angles at C and F be supposed not to be less than a right angle: I say again that also thus the triangle ABC is equiangular to the triangle DEF.

For the fame things being constructed, we shall demonftrate in like manner that BC is equal to BG; so that also (by 5. 1.) the angle at C is equal to the angle BGC: but the angle at C is not less than a right angle; therefore neither is the angle BGC lefs than a right angle; therefore the two angles of the triangle BGC are not less than two right angles, which (by 17. 1.) is impoffible: Therefore again the angle ABC is not unequal to the angle DEF; therefore it is equal: but also (by fupp.) the angle at A is equal to the angle at D; therefore (by 32. 1.) the remaining angle at C is equal to the remaining angle at F: wherefore the triangle ABC is equiangular to the triangle DEF.

Wherefore if two triangles have one angle equal to one angle; but the fides about two other angles proportionals; and each of the remaining angles, at the fame time, either lefs or not less than a right angle: the triangles will be equiangular and will have the angles equal about which the fides are, which are proportionals. Which was to be demonftrated.

PROP.

If in a right angled triangle, a perpendicular be drawn from the right angle to the bafe; the triangles at the perpendicular are fimilar to the whole triangle and to one another.

Let ABC be a right angled triangle, having the angle BAC a right angle; and let AD be drawn from the point A perpendicular to BC: I say that each of the triangles ABD, ADC is fimilar to the whole ABC, and alfo to one another.

C

A

D B

For fince 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, both ABC and ABD; therefore (by 32. 1.) the remaining angle ACB is equal to the remaining angle BAD; therefore the triangle ABC is equiangular to the triangle ABD; wherefore it is, (by 4. 6.) as BC fubtending the right angle of the triangle ABC is to BA fubtending the right angle of the triangle ABD, fo is the fame AB fubtending the angle at C of the triangle ABC to BD fubtending the angle equal to that at C, viz. the angle BAD of the triangle ABD and moreover fo is AC to AD fubtending the angle at B common to the two triangles: wherefore the triangle ABC is equiangular to the triangle ABD; and has the fides about the equal angles proportionals: therefore (by 1. def. 6.) the triangle ABC is fimilar to the triangle ABD. Certainly in the fame manner we shall demonftrate, that the triangle ADC is also fimilar to the triangle ABC; therefore each of the triangles ABD, ADC is fimilar to the whole triangle ABC.

I fay also that the triangles ABD, ADC are fimilar to one another. For because the right angle BDA is equal to the right angle ADC but the angle BAD has been demonstrated to be equal to the angle at C; therefore alfo the remaining angle at B is equal to the remaining angle DAC: wherefore the triangle ABD is equiangular to the triangle ADC: therefore it is, as BD fubtending the angle BAD of the triangle ABD is to DA fubtending the angle at C, of the triangle ADC, equal to the angle BAD fo is the fame AD fubtending the angle at B of the triangle ABD, to DC subtending

*E 2

Book VI.

Book VI. tending the angle DAC of the triangle ADC, equal to the angle at B and befides fo is BA fubtending the right angle ADB to AC fubtending the right angle ADC: therefore (by 1. def. 6.) the triangle ABD is fimilar to the triangle ADC.

Wherefore if in a right angled triangle, a perpendicular be drawn from the right angle to the base; the triangles at the perpendicular are fimilar to the whole triangle and to one another. Which was to be demonstrated.

Cor. Certainly it is manifeft from this, that if a perpendicular be drawn from the right angle in a right angled triangle to the bafe; the perpendicular drawn is a mean proportional between the segments of the base: and befides the fide adjacent to the fegment is a mean proportional between the bafe andanyo ne of the fegments.

PROP. IX.

To cut off any part required from a given ftraight line.

Let AB be the given straight line: it is required to cut off any part from AB.

Let a third part be required; and let any straight line AC be drawn from the point A, containing any angle, which may accidentally happen, with the line AB; and let D be taken in AC any point which may accidentally happen; and (by 3. 1.) let DE, EC be made equal to AD; and let BC be joined; and through the point D, let DF be drawn (by 31. 1.) parallel to BC.

Wherefore becaufe DF has been drawn parallel to BC one of the fides of the triangle ABC; therefore (by 2. 6.) there is this proportion, as CD is to DA fo is BF to FA: but CD is the double of DA; therefore BF is the double of FA; therefore BA is the triple of AF.

Therefore the required third part AF hath been cut off from the given ftraight line AB. Which was to be done,

F

A

D

E

B

PROP.

To cut an undivided straight line, in like manner as a given straight line hath been cut.

Let AB be the given undivided ftraight line, and AC the one which hath been cut it is required to cut the undivided straight line AB in like manner as AC has been cut.

Let AC be cut in the points D, E: and let the lines be placed fo as to contain any angle which may accidentally happen; and let BC be joined; and let DF, GE be drawn (by 31. 1.) through the points D, E parallel to BC: and through the point D, let DHK be drawn parallel to AB.

B

A

F

D

G

H

E

K

C

Therefore each of the figures FH, HB is a parallelogram therefore (by 34. 1.) DH is equal to FG; and HK to GB: And because HE hath been drawn parallel to KC one of the fides of the triangle DKC; (by 2.6.) there is this proportion, as CE is to ED fo is KH to HD; but KH is equal to BG, and HD to GF; therefore it is (by 7. 5.) as CE is to ED fo is BG to GF: Again, because FD hath been drawn parallel to EG one of the fides of the triangle AGE; therefore (by 2.6.) there is this proportion, as ED is to DA fo is GF to FA; but it has been alfo demonftrated that as CE is to ED fo is BG to GF It is 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 straight line AB hath been cut in like manner as the given straight line AC had been cut. Which was to be done.

To find a third proportional to two given straight lines.

Let AB, AC be the two given straight lines; and let them be placed containing any angle which may accidentally happen: it is required to fiind a third proportional to AB, AC.

Book VI.

For let AB, AC be produced to the points D, E; and let BD be made equal to AC, and let BC be joined; and through the point

D,