Book VI.

See N.

a 32. I.

IN

PROP. VIII. THEOR.

Na right angled triangle, if a perpendicular be drawn from the right angle to the bafe; the triangles on each fide of it are fimilar to the whole triangle, and to one another.

Let ABC be a right angled triangle, having the right angle BAC; and from the point A let AD be drawn perpendicular to the base BC: The triangles ABD, ADC are fimilar to the whole triangle ABC, and to one another.

A

Because the angle BAC is equal to the angle ADB, each of them being a right angle, and that the angle at B is common to the two triangles ABC, ABD; the remaining angle ACB is equal to the remaining angle BAD: Therefore the triangle ABC is equiangular to the triangle ABD, and the fides about their equal angles B are proportionals b; wherefore c 1. def. 6. the triangles are fimilar: In

b 4. 6.

D C

the like manner it may be demonftrated, that the triangle ADC is equiangular and fimilar to the triangle ABC: And the triangles ABD, ADC, being both equiangular and fimilar to ABC, are equiangular and fimilar to each other. Therefore, in a right angled, &c. Q. E. D.

COR. From this it is manifeft, that the perpendicular drawn from the right angle of a right angled triangle to the base, is a mean proportional between the fegments of the bafe: And alfo, that each of the fides is a mean proportional between the bafe, and its fegment adjacent to that fide: Because in the triangles BDA, ADC, BD is to DA, as DA to DC b; and in the triangles ABC, DBA, BC is to BA, as BA to BD b; and in the triangles ABC, ACD, BC is to CA, as CA to CD b.

PROP.

Book VI

PRO P. IX. PRO B.

ROM a given ftraight line to cut off any part re- See N.

Fquired.

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

From the point A draw a straight line AC making any angle with AB; and in AC take any point D, and take AC the fame multiple of AD, that AB is of the part which is to be cut off from it; join BC, and draw DE parallel to it: Then AE is the part required to be cut off.

Because ED is parallel to one of the fides E of the triangle ABC, viz. to BC, as CD is to DA, fo is a BE to EA; and, by compofition b, CA is to AD, as BA to AE: But CA is a multiple of AD; therefore ‹ BA is the fame multiple of AE: Whatever part

therefore AD is of AC, AE is the fame B

part of AB: Wherefore, from the straight

A

D

C

line AB the part required is cut off. Which was to be done.

PROP. X. PROB.

O divide a given ftraight line fimilarly to a given divided straight line, that is, into parts that shall have the fame ratios to one another which the parts of the divided given ftraight line have.

Let AB be the straight line given to be divided, and AC the divided line; it is required to divide AB fimilarly to AC.

a 2. 6. b 18.5. c D. 5.

Let AC be divided in the points D, E; and let AB, AC be placed fo as to contain any angle, and join BC, and through the points D, E, draw a DF, EG parallels to it; and through D a 31. 1. draw DHK parallel to AB: Therefore each of the figures FH, HB, is a parallelogram; wherefore DH is equal to FG, and b 34. X.

L 4

·HK

c 2.6.

F

Book VI. HK to GB: And because HE is pa-
rallel to KC, one of the fides of the
triangle DKC, as CE to ED, fo is
c KH to HD: But KH is equal to
BG, and HD to GF; therefore, as
CE to ED, fo is BG to GF: Again, G
because FD is parallel to EG, one of
the fides of the triangle AGE, as ED B
to DA, fo is GF to FA: But it has
been proved that CE is to ED, as

BG to GF; and as ED to DA, fo GF to FA: Therefore the given ftraight line AB is divided fimilarly to AC. Which was

to be done.

a 31. r.

b 2.6.

Let AB, AC be the two given ftraight lines, and let them
be placed fo as to contain any angle; it is
required to find a third proportional to AB,
AC.

Produce AB, AC to the points D, E;
and make BD equal to AC; and having B
joined BC, through D, draw DE parallel to
it a.

A

C

D

E

Because BC is parallel to DE, a fide of the triangle ADE, AB is b to BD, as AC to CE: But BD is equal to AC; as therefore AB to AC, fo is AC to CE. Wherefore to the two given ftraight lines AB, AC a third proportional CE is found. Which was to be done.

Let A, B, C be the three given straight lines; it is required to find a fourth proportional to A, B, C.

Take

Take two ftraight lines DE, DF, containing any angle EDF; Book VI.

therefore, as A is to B, fo is C to HF. Wherefore to the three given ftraight lines, A, B, C a fourth proportional HF is found. Which was to be done.

Let AB, BC be the two given straight lines; it is required to find a mean proportional between them.

Place AB, BC in a straight line, and upon AC describe the

femicircle ADC, and from the point B draw a BD at right angles to AC, and join AD, DC.

Because the angle ADC in a femicircle is a right angle b, and because in the right angled triangle ADC, DB is drawn from

the right angle perpendicular to A

the base, DB is a mean propor

D

B C

a II. I.

b 31. 3.

tional between AB, BC the fegments of the bafe : Therefore c Cor. 8. 6. between the two given straight lines AB, BC, a mean proportional DB is found. Which was to be done.

PROP.

Book VI.

a 14. I.

b 7.5. C I. 6.

d II. 5.

€9.5.

PROP. XIV. THE OR.

EQUAL parallelograms which have one angle of the

one equal to one angle of the other, have their fides about the equal angles reciprocally proportional: And parallelograms that have one angle of the one equal to one angle of the other, and their fides about the equal angles reciprocally proportional, are equal to one

another.

Let AB, BC be equal parallelograms, which have the angles at B equal, and let the fides DB, BE be placed in the fame ftraight line; wherefore alfo FB, BG are in one straight line a; The fides of the parallelograms AB, BC about the equal angles, are reciprocally proportional; that is, DB is to BE, as GB to BF.

A

F

E

D

B

Complete the parallelogram FE; and because the parallelogram AB is equal to BC, and that FE is another parallelogram, AB is to FE, as BC to FE b: But as AB to FE, fo is the bafe DB to BE c; and, as BC to FE, fo is the bafe GB to BF; therefore, as DB to BE, fo is GB to BF d. Wherefore, the fides of the parallelograms AB, BC about their equal angles are reciprocally proportional.

G

But, let the fides about the equal angles be reciprocally proportional, viz. as DB to BE, fo GB to BF; the parallelogram AB is equal to the parallelogram BC.

Because, as DB to BE, fo is GB to BF; and as DB to BE, fo is the parallelogram AB to the parallelogram FE; and as GB to BF, fo is the parallelogram BC to the parallelogram FE; therefore as AB to FE, fo BC to FE d: Wherefore the parallelogram AB is equal e to the parallelogram BC. Therefore equal parallelograms, &c. Q. E. D.

PROP.