Book VI. equal to the angle GEF, and the triangle DEF equal to the triangle GEF, and the remaining angles of the one triangle equal to the remaining angles of the other, which are oppofite to the equal fides. Therefore the angle DFE is equal to the angle GFE, and the angle EDF equal to the angle EGF. And because the angle DEF is equal to the angle GEF, and [by conftruction] the angle GEF is equal to the angle ABC; the angle ABC will be equal to the angle DEF. By the fame reafon the angle ACB is equal to the angle DFE, and also the angle A equal to the angle D: Therefore the triangles ABC, DEF are equiangular.

If therefore two triangles have their fides proportional, the triangles will be equiangular: and those angles will be equal which are oppofite to the homologous fides. Which was to be demonftrated.

PRO P. VI. THEOR.

If two triangles have one angle of the one equal to one angle of the other, and the fides about the equal angles proportional: thefe triangles will be equiangular, having thofe angles equal which are oppofite to the homologous fides.

Let two triangles ABC, DEF have one angle BAC of the one, equal to one angle EDF of the other, and the fides about these equal angles proportional, viz. as BA is to A c, fo is ED to DF: I fay the triangles ABC, DEF are equiangular, the angle ABC being equal to the angle DEF, and the angle ACB equal to the angle DFE.

For [by 23. 1.] at the right line DF, and at the points D, F in it, make the angle FDG equal to BAC or EDF, and the angle D F G equal to the angle ACB.

Then will the remaining angle at B [by 32. 1.] be equal to the remaining angle at G. Therefore the triangles ABC,

DGF are equiangular : and fo [by 4. 6.] as BA Gis to AC, fo is G D to D F. But [by fuppofition] as BA is to AC, fo is ED to DF: Therefore as ED is to D F, fo is [by 11. 5.]

GD

GD to DF: Wherefore [by 9. 5.] ED is equal to DG, and DF is common: Therefore the two fides E D, DF are equal to the two fides GD, DF, and the angle EDF is equal to the angle GDF: Wherefore the bafc EF [by 4. 1.] is equal to the base FG, and the triangle D E F equal to the triangle GDF, and the remaining angles equal to the remaining angles, each to each, which are oppofite to the equal fides: Therefore the angle D F G is equal to the angle DFE; and the angle at G equal to the angle at E. But the angle DF G [by construction] is equal to the angle ACB: Therefore the angle ACB is equal to the angle DFE. But B A C is supposed to be equal to the angle EDF: Therefore [by 32. 1.] the remaining angle at B is equal to the remaining angle at E: Wherefore the triangles A BC, D E F are equiangular.

Therefore if two triangles have one angle of the one equal to one angle of the other, and the fides about the equal angles proportional; these triangles will be equiangular, having those angles equal which are oppofite to the homologous fides, Which was to be demonstrated.

If two triangles have one angle of the one equal to one angle of the other, and the fides about two other angles proportional, and have alfo each of the remaining angles either lefs or not less than a right angle: thofe triangles will be equiangular, and have thofe angles equal about which are the proportional fides.

And

Let the two triangles be ABC, DEF, having one angle BAC of the one, equal to one angle EDF of the other, and the fides about other angles ABC, DEF proportional, fo that DE be to EF, as A B is to BC. having also each of the remaining angles at c, F firft less than a right angle: I say the triangles A BC, DEF will be equiangular, the angle ABC equal to the angle DEF, and the remaining angle at c equal to the remaining angle

at F.

For if the angle ABC be unequal to the angle DEF, one of them will be the greater, which let be ABC. And

at

at the right line AB and at the point в in it make [by 23. 1.] the angle ABG equal to the angle DE F.

Now because the angle A is equal to the angle D, and the angle ABG equal to the angle DEF; [by 32. 1.] the

D

AA

B

G

F

remaining angle AGB will be equal to the remaining angle DFE: Therefore the triangle ABG is equiangular to the triangle D E F Wherefore [by 4. 6.] as A B is to BG, fo is DE to EF. But as DE is to EF, fo is [by fuppofition] A B to BC: Therefore as A B is to BC, fo [by 11. 5.] is A B to BG; and fo A B has the fame ratio to BC or BG: Wherefore BC is equal to BG; and accordingly [by 5. 1.] the angle BGC is equal to the angle BCG. But the angle at c is supposed to be lefs than a right angle: Therefore BGC will also be less than a right angle, wherefore, [by 13. 1.] the adjacent angle AGB will be greater than a right angle. But it has been proved that the angle AGB is equal to the angle at F: Therefore the angle at F is greater than a right angle. But it is fuppofed to be less than a right angle; which is abfurd: Therefore the angle ABC is not unequal to the angle DEF; they are therefore equal. But the angle at A is equal to the angle at D; wherefore the remaining angle at c is equal to the remaining angle at F: Therefore the triangles ABC, DEF are equiangular.

Again, let us fuppofe each of the angles c, F to be not lefs than a right angle: I fay moreover that the triangles ABC, DEF will be equiangular.

For the fame construction remaining, we demonftrate after the like manner that BC is equal to BG, and the angle c equal to the angle BGC. But the angle at c is not lefs than a right angle: Therefore BGC is not less than a right angle: Wherefore the two angles of the triangle BGC are not lefs than two right angles, which [by 17. 1.] is impoffible: Wherefore again the angle ABC is not unequal to the angle DEF: it is therefore equal to it. But the angle at A is equal to the angle at p: Therefore the remaining angle at c [by 32. 1.] is equal to the remaining angle at F and accordingly the triangles ABC, DEF are equiangular.

If therefore two triangles have one angle of the one equal to one angle of the other, and the fides about two other angles proportional, and have also each of the remaining angles either lefs or greater than a right angle: these triangles will be equiangular, and fhall have those angles equal, about which the fides are proportional. Which was to be demonftrated.

e Euclid has added that each of the remaining angles B, E must be either both less or both not less than a right angle, for otherwife, the whole hypothefis remaining, it would not follow that the triangles would be equiangular. For if in the fame triangle ABC, BG were equal to в C, which would happen when the angle B is acute, and greater than a: In this cafe the two triangles ABG, DEF would have one, angle of the one equal to one angle of the other, viz. the angle A and the angle D, and the fides about the other angles ABG, DEF proportional, that is, as A B is to B G, fo is DE to EF, and yet the triangles ABG, DEF would not be equiangular; and that because one angle AGB is greater than a right angle, and the other E F D lefs.

Taquet fays this propofition is fcarcely of any ufe. But herein he is more prompted by dislike than certainty.

THEOR.

PRO P. VIII. If a perpendicular be drawn from the right angle to ibe bafe of a right angled triangle: this perpendicular will divide that right angled triangle into two other triangles; each of which will be fimilar to one another, and to the whole right angled triangle.

Let there be a right angled triangle ABC, having the right angle B A C ; and let AD be the perpendicular drawn from the point A to the base BC: I fay the triangles A BD, ADC are fimilar to one another, and alfo to the whole triangle ABC.

For because the angle B A C is equal to the angle A DB, each of them being a right angle, and the angle which is at в is common to both the triangles ABC, A BD; the remaining angle ACB [by 32. 1.] will be equal to the remaining angle BAD: Therefore the triangle ABC is equiangular to the triangle A B.D. Wherefore [by 4. 6.] as

the

A

D

the fide BC oppofite to the right angle of the triangle ABC, is to the fide BA, oppofite to the right angle of the triangle ABD, fo is the fide A B oppofite to the angle at c of the triangle ABC to the fide BD oppofite to an angle equal to the angle at c, viz. BAD of the triangle ABD. And fo alfo is AC to the fide A D oppofite to the angle at B, which is common to both the triangles: Therefore the triangle ABC is equiangular to the triangle ABD, and has the fides about the equal angles proportional: therefore [by 1. def. 6.] the triangle ABC is fimilar to the triangle A B D. By the fame reason we demonstrate that the triangle ADC is alfo fimilar to the triangle ABC; wherefore the triangles ABD, ADC are each fimilar to the whole triangle A BC. I fay moreover that the triangles ABD, ADC are also fimilar to one another.

For because the right angle BDA is equal to the right angle A DC; but B A D has been proved to be equal to the angle at c; [by 32. 1.] the remaining angle at в will be equal to the remaining angle DAC: Therefore the triangle A B D is equiangular to the triangle ADC. Wherefore [by 4. 6.] as the fide BD of the triangle ABD which is oppofite to the angle B AD, is to the fide DA of the triangle ADC, which is oppofite to the angle at c, equal to the angle BAD, fo is the fide A D of the triangle A B D, which is oppofite to the angle at B, to the fide DC of the triangle ADC, which is oppofite to the angle DAC, being equal to the angle at B. And fo alfo is the fide B A oppofite to the right angle A D B, to the fide AC which is oppofite to the right angle ADC: Therefore [by 1. def. 6.] the triangle ABD is fimilar to the triangle ADC.

Wherefore if a perpendicular be drawn from the right angle to the base of a right angled triangle; this perpendicular will divide that right angled triangle into two other triangles, each of which will be fimilar to one another, and to the whole right angled triangle. Which was to be demonftrated.

Corollary. From hence it is manifeft, that the perpendicular drawn from the right angle to the base of a right an

gled