triangle ABC; therefore, ex aequali, as twice BD is to the half of DA, that is, as quadruple of BD is to DA, that is, as quadruple of FH to HE, fo is twice the rectangle DB, BC to the triangle ABC.

PROP. LXXV.

IF a triangle has a given acute angle, the space by which the fquare of the fide fubtending the acute angle is less than the fquares of the fides which contain it, fhall have a given ratio to the triangle.

65

Let the triangle ABC have a given acute angle ABC, and draw AD perpendicular to BC, the space by which the fquare of AC is less than the fquares of AB, BC, that is, the double a 13. 2. of the rectangle contained by CB, BD, has a given ratio to the triangle ABC.

A

Because the angles ABD, ADB are each of them given, the triangle ABD is given in fpecies; and therefore the ratio of BD to DA is given: And as BD to DA, fo is the rectangle CB, BD to the rectangle CB, AD; therefore the ratio of thefe rectangles is given, as alfo the ratio of twice the rectangle CB, BD to the rectangle CB, AD; but the rectangle CB, AD has a given ratio to its half the triangle ABC; therefore the B ratio of twice the rectangle CB, BD to the triangle ABC is given; and twice the rectangle CB, BD is the space by which the fquare of AC is lefs than the fquares of AB, BC; there. fore the ratio of this fpace to the triangle ABC is given: And the ratio may be found as in the preceding propofition.

IF

DC b 9. daf.

F from the vertex A of an ifofceles triangle ABC, any straight line AD be drawn to the bafe BC, the fquare of the fide AB is equal to the rectangle BD, DC of the fegments of the bafe together with the fquare of AD; but if AD be drawn to the bafe produced, the fquare of AD is equal to the rectangle BD, DC together with the fquare of AB.

a

CAS. 1. Bifect the bafe BC in E, and join AE which will be perpendicular to BC; wherefore the fquare of AB is equal to the fquares of AE, EB; but the fquare of EB is equal to the rectangle BD, DC together with the fquare of DE; there

A

a 8. 1.

b 4. 7. I c 5. I.

fore the fquare of AB is equal to the D BDE C

fquares

b 47. I.

67.

b 43. dat.

€ 50. dat.

d 1. 6.

fquares of AE, ED, that is, to the fquare of AD, together with the rectangle BD, DC; the other cafe is shown in the fame way by 6. 2. Elem.

I'

PROP. LXXVI.

F a triangle have a given angle, the excefs of the square of the straight line which is equal to the two fides that contain the given angle, above the square of the third fide, fhall have a given ratio to the triangle.

Let the triangle ABC have the given angle BAC, the excefs of the fquare of the ftraight line which is equal to BA, AC together above the fquare of BC, shall have a given ratio to the triangle ABC.

Produce BA, and take AD equal to AC, join DC and produce it to E, and through the point B draw BE parallel to AC; join AE, and draw AF perpendicular to DC; and becaufe AD is equal to AC, BD is equal to BE; and BC is drawn from the vertex B of the ifofceles triangle DBE, therefore, by the Lemma, the fquare of BD, that is, of BA and AC together, is equal to the rectangle DC, CE together with the square of BC; and, therefore, the fquare of BA, AC together, that is, of BD, is greater than the fquare of BC by the rectangle DC, CE; and this rectangle has a given

ratio to the triangle ABC: Because

the angle BAC is given, the adjacent B
angle CAD is given; and each of the
angles ADC, DCA is given, for each

a

D

C

E

a5.& 32. 1. of them is the half of the given angle BAC; therefore the triangle ADC is given in fpecies; and AF is drawn from its vertex to the base in a given angle; wherefore the ratio of AF to the base CD is given; and as CD to AF, fo is a the rectangle DC, CE to the rectangle AF, CE; and the ratio of the rectangle AF, CE to its half the triangle ACE is given; therefore the ratio of the rectangle DC, CE to the triangle ACE, that is, to the triangle ABC, is given; and the rectangle DC, CE is the excefs of the fquare of BA, AC together above the square of BC; therefore the ratio of this excess to the triangle ABC is given.

€ 41. I.

f 37. I.

9. dat.

The ratio which the rectangle DC, CE has to the triangle ABC is found thus: Take the ftraight line GH given in poli

tion and magnitude, and at the point G in GH make the angle HGK equal to the given angle CAD, and take GK equal to GH, join KH, and draw GL perpendicular to it: Then the ratio of HK to the half of GL is the fame with the ratio of the rectangle DC, CE to the triangle ABC: Because the angles HGK, DẶC at the vertices of the ifofceles triangles GHK, ADC are equal to one another, these triangles are fimilar; and becaufe GL, AF are perpendicular to the bafes HK, DC, as 24. 6. HK to GL, fo is (DC to AF, and fo is) the rectangle DC, CE to the rectangle AF, CE; but as GL to its half, fo is the rectangle AF, CE to its half, which is the triangle ACE, or the triangle ABC; therefore, ex aequali, HK is to the half of the ftraight line GL, as the rectangle DC, CE is to the triangle ABC.

h

COR. And if a triangle have a given angle, the space by which the fquare of the straight line which is the difference of the fides which contain the given angle is lefs than the fquare of the third fide, fhall have a given ratio to the triangle. This is demonstrated the fame way as the preceding propofition, by help of the fecond cafe of the Lemma.

PROP. LXXVII.

h

22. 5.

I.

IF the perpendicular drawn from a given angle of a See N. triangle to the oppofite fide, or base, has a given ra

tio to the base, the triangle is given in species.

Let the triangle ABC have the given angle BAC, and let the the perpendicular AD drawn to the bafe BC, have a given ratio to it, the triangle ABC is given in species.

If ABC be an ifofceles triangle, it is evident that if any a 5.& 32.1.

one of its angles be given, the reft are aifo given; and therefore the triangle is given in fpecies, without the confideration of the ratio of the perpendicular to the bafe, which in this cafe is given by prop. 50.

But when ABC is not an ifofceles triangle, take any straight line EF given in pofition and magnitude, and upon it defcribe

the

1

b 2. dat.

€ 30. dat.

d 31. dat.

the fegment of a circle EGF containing an angle equal to the given angle BAC, draw GH bifecting EF at right angles, and join EG, GF: Then, fince the angle EGF is equal to the angle BAC, and that EGF is an ifofceles triangle, and ABC is not, the angle FEG is not equal to the angle CBA: Draw EL making the angle FEL equal to the angle CBA; join FL, and draw LM perpendicular to EF; then, because the triangles ELF, BAC are equiangular, as alfo are the triangles MLE, DAB, as ML to LE, fo is DA to AB; and as LE to EF, fo is AB to BC; wherefore, ex aequali, as LM to EF, fo is AD to BC; and because the ratio of AD to BC is given, therefore the ratio of LM to EF is given; and EF is given, wherefore LM alfo is given. Complete the parallelogram LMFK; and because LM is given, FK is given in magnitude; it is alfo given in pofition, and the point F is given, and confequently the point K; and becaufe through K the straight line KL is drawn parallel to EF which is given in pofition, therefore KL is given in pofition;

с

e 28. dat.

f 29. dat. 8 42. dat.

B RDC

E OHM F

and the circumference ELF is given in pofition; therefore the point L is given. And because the points L, E, F are given, the ftraight lines LE, EF, FL are given f in magnitude; therefore the triangle LEF is given in fpecies; and the triangle ABC is fimilar to LEF, wherefore also ABC is given in fpecies.

Because LM is lefs than GH, the ratio of LM to EF, that is, the given ratio of AD to BC, must be less than the ratio of GH to EF, which the ftraight line, in a fegment of a circle containing an angle equal to the given angle, that bifects the base of the fegment at right angles, has unto the bafe.

COR. I. If two triangles, ABC, LEF have one angle BAC equal to one angle ELF, and if the perpendicular AD be to the bafe BC, as the perpendicular LM to the base EF, the triangles ABC, LEF are fimilar.

Defcribe the circle EGF about the triangle ELF, and draw LN parallel to EF, join EN, NF, and draw NO perpendicu lar to EF; because the angles ENF, ELF are equal, and that

the

the angle EFN is equal to the alternate angle FNL, that is, to the angle FEL in the fame fegment; therefore the triangle NEF is fimilar to LEF; and in the fegment EGF there can be no other triangle upon the bafe EF, which has the ratio of its perpendicular to that bafe the fame with the ratio of LM or NO to EF, because the perpendicular must be greater or lefs than LM or NO; but, as has been fhewn in the preceding demonftration, a triangle fimilar to ABC can be defcribed in the fegment EGF upon the bafe EF, and the ratio of its perpendicular to the base is the fame, as was there fhewn, with the ratio of AD to BC, that is, of LM to EF; therefore that triangle must be either LEF, or NEF, which therefore are fimilar to the triangle ABC.

COR. 2. If a triangle ABC has a given angle BAC, and if the ftraight line AR drawn from the given angle to the oppofite fide BC, in a given angle ARC, has a given ratio to BC, the triangle ABC is given in fpecies.

h

Draw AD perpendicular to BC; therefore the triangle ARD is given in fpecies; wherefore the ratio of AD to AR is given; and the ratio of AR to BC is given, and confequently the ra- h 9. dat. tio of AD to BC is given; and the triangle ABC is therefore given in fpecies i.

COR. 3. If two triangles ABC, LEF have one angle BAC equal to one angle ELF, and if straight lines drawn from these angles to the bases, making with them given and equal angles, have the fame ratio to the bases, each to each; then the triangles are fimilar; for having drawn perpendiculars to the bases from the equal angles, as one perpendicular is to its base, so is the other to its bafe ; wherefore, by Cor. 1. the triangles are

fimilar.

A triangle fimilar to ABC may be found thus: Having defcribed the fegment EGF and drawn the ftraight line GH as was directed in the propofition, find FK which has to EF the given ratio of AD to BC; and place FK at right angles to EF from the point F; then becaufe, as has been fhewn, the ratio of AD to BC, that is, of FK to EF, must be lefs than the ratio of GH to EF; therefore FK is lefs than GH; and confequently the parallel to EF drawn through the point K, muft meet the circumference of the fegment in two points: Let L be either of them, and join EL, LF, and draw LM perpendicular to EF; then, becaufe the angle BAC is equal to the angle ELF, and that AD is to BC, as KF, that is, LM to EF, the triangle ABC is fimilar to the triangle LEF, by Cor. 1.

PROB

i 77. dat,

k

54. 6.

22. 5.