gular and similar to the triangle ABC: and the triangles ABD, ADC, being each equiangular and similar to ABC, are equiangular and similar to one another, Therefore, in a right angled, &c. Q. E. D.

COR. From this it is manifest, that the perpendicular drawn, from the right angle of a right angled-triangle, to the base, is a mean proportional between the segments of the base; and also that each of the sides is a mean proportional between the base, and its segment adjacent to that side. For in the triangles BDA, ADC,

BD : DA☀☀ DA : DC (4, 6.); and in the triangles ABC, BDA, BC: BA BA: BD (4. 6.); and in the triangles ABC, ACD, BC CA;: CA: CD (4. 6.).

+ PROP. IX. PROB.

From a given straight line to cut off any part required, that is, a part which shall be contained in it a given number of times.

Let AB be the given straight line; it is required to cut off from AB, a part which shall be contained in it a given number of times. From the point A draw a straight line AC making any angle with AB; and in AC take any point D, and take AC

such that it shall contain AD, as oft as AB is to contain 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 sides of the triangle ABC, viz. to BC, CD: DA :: DE: EA (2. 6.); and by composition (18. 5.), CA: AD :: BA AE: But CA is a multiple of AD; therefore (C. 5.) BA is the same multiple of AE, or contains AE the same num- B ber of times that AC contains AD; and there

fore, whatever part AD is of AC, AE is the same of AB; wherefore, from the straight line AB the part required is cut off. Which was to be done.

To divide a given straight line similarly to a given divided straight line, that is, into parts that shall have the same ratios to one another which the parts of the divided given straight line have.

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

Let AC be divided into the points D, E; and let AB, AC be placed so as to contain any angle, and join BC, and through the points D, E, draw (31. 1.) DF, EG, parallel to BC; and through D draw DHK, parallel to AB; therefore each of the figures FH, HB, is a parallelo

gram: wherefore DH is equal (34. 1.) to FG, and HK to GB: and because HE is parallel to KC, one of the sides of the triangle DKC, CE: ED :: (2. 6.) KH : HD: But KH=BG, and HD=GF; therefore CE: ED: BG: GF: Again, because FD is parallel to EG, one of the sides of the triangle AGE, ED: DA: GF: FA: But it has been proved that CE: ED BG: GF; therefore the given straight line AB is divided similarly to AC. Which was to be done.

PROP. XI. PROB.

To find a third proportional to two given straight lines. Let AB, AC be the two given straight lines, and let them be placed so as to contain any angle; it is requir

ed to find a third proportional to AB

AC.

Produce AB, AC to the points D,E; and make BD equal to AC; and having joined BC, through D draw DE parallel to it (31. 1.).

Because BC is parallel to DE, a side of the triangle ADE, AB: (2. 6.) BD AC: CE; but BE=AC: therefore AB: AC:: AC: CE. Wherefore to the two

B

A

given straight lines AB, AC a third proportional, CE is found. Which was to be done.

PROP. XII. PROB.

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

and

Take two straight lines DE, DF, containing any angle EDF; upon these make DG equal to A, GE equal to B and DH equal to C; and having joined GH, draw EF parallel (31. 1.) to it through the

D

A.

B

H

point E. And because GH is parallel to EF, one of the sides of the triangle DEF, DG: GE :: DH : HF (2. 6.); but DG=A, GE=B, and DH C; and therefore A: BC: HF. Wherefore to the three given straight lines, A, B, C, a fourth proportional HF is found. Which was to be done.

PROP. XIII. PROB.

To find a mean proportional between two given straight lines.

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

D

Place AB, BC in a straight line, and upon AC describe the semicir ele ADC, and from the point B (11. 1.) draw BD at right angles to AC, and join AD, DC.

Because the angle ADC in a semicircle is a right angle (31. 3.) 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

B

tional between AB, BC, the segments of the base (Cor. 8. 6.); therefore between the two given straight lines AB, BC, à mean proportional DB is found. Which was to be done.

PROP. XIV. THEOR.

Equal parallelograms which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: And parallelograms which have one angle of the one equal to one angle of the other, and their sides 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 sides DB, BE be placed in the same straight line; wherefore also FB, BG are in one straight line (14. 1.): the sides of the parallelograms, AB, BC, about the equal angles, are reciprocally proportional; that is, DB is to BE, as GB to BF.

Complete the parallelogram FE; and because the parallelograms AB, BC are equal, and FE is another parallelogram,

AB: FE BC: FE (7. 5.):

but because the parallelograms AB, FE have the same altitude, AB : FE :: DB : BE (1. 6.), also,

BC: FE:: GB: BF (1. 6.); therefore

DB: BE:: GB: BF (11. 5.). Wherefore, the sides of the parallelograms AB, BC about their equal angles are reciproeally proportional.

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

Because, DB: BE :: GB : BF, and DB: BE :: AB: FE, and GB: BF BC: EF, therefore, AB FE :: BC: FE (11. 5.); Wherefore the parallelogram AB is equal (9.5.) to the parallelogram BC. Therefore equal parallelograms, &c. Q. E. D.

PROP. XV. THEOR.

Equal triangles which have one angle of the one equal to one angle of the other have their sides about the equal angles reciprocally proportional: And triangles which have one angle in the one equal to one angle in the other, and their sides about the equal angles reciprocally proportional, are equal to one another.

Let ABC, ADE be equal triangles, which have the angle BAC equal to the angle DAE; the sides about the equal angles of the triangles are reciprocally proportional; that is, CA is to AD, as EA to AB.

Let the triangles be placed so that their sides CA, AD be in one straight line; wherefore also EA and AB are in one straight line (14.

B

A

E

1.); join BD. Because the triangle ABC is equal to the triangle ADË, and ABD is another triangle; therefore, triangle CAB : triangle BAD: triangle EAD: triangle BAD; but CAB: BAD :: CA: AD and EAD : BAD :: EA: AB; therefore CA : AD :: EA : AB (11. 5.), wherefore the sides of the triangles ABC, ADE about the equal angles are reciprocally proportional.

But let the sides of the triangles ABC, ADE, about the equal angles be reciprocally proportional, viz. CA to AD, as EA to AB; the triangle ABC is equal to the triangle ADE.

Having joined BD as before; because CA: AD:: EA: AB; and since CA: AD: : triangle ABC triangle BAD (1. 6.); and also EA AB triangle EAD: triangle BAD (11. 5.); therefore, triangle ABC triangle BAD: : triangle EAD: triangle BAD; that is,

:

the triangles ABC, EAD have the same ratio to the triangle BAD, wherefore the triangle ABC is equal (9. 5.) to the triangle EAD. · Therefore equal triangles, &c. Q. E. D.

PROP. XVI. THEOR.

If four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means; And if the rectangle contained by the extremes be equal to the rectangle contained by means, the four straight lines are proportionals.

Let the four straight lines, AB, CD, E, F be proportionals, viz. as AB to CD, so E to F; the rectangle contained by AB, F is equal to the rectangle contained by CD, E.

From the points A, C draw (11.1.) AG, CH at right angles to AB, CD; and make AG equal to F, and CH equal to E, and complete the parallelograms BG, DH. Because AB: CD: E: F; and since E-CH, and F=AG, AB: CD (7. 5.) :: CH: AG: therefore the sides of the parallelograms BG, DH about the equal angles are reciprocally proportional; but parallelograms which have their sides about equal angles reciprocally proportional, are equal to one another (14. 6.); therefore the parallelogram BG is equal to the parallelogram DH: and the parallelogram BG is E. contained by the straight lines AB, F. F; because AG is equal to F; and the parallelogram DH is contained by CD and E, because CH is equal to E: therefore the rectangle contained by the straight lines AB, F is equal to that which is contained by CD and E.

G

A

H

B

C

D

And if the rectangle contained by the straight lines AB, F be equal to that which is contained by CD, E; these four lines are proportionals, viz. AB is to CD, as E to F. The same construction being made, because the rectangle contained by the straight lines AB, F is equal to that which is contained by CD, E, and the rectangle BG is contained by AB, F, because AG is equal to F; and the rectangle DH, by CD, E, because CH is equal to E; therefore the parallelogram BG is equal to the parallelogram DH and they are equiangular: but the sides about the equal angles of equal parallelograms are reciprocally proportional (14. 6.): wherefore AB: CD: CH: AG; but CH E, and AG=F, therefore AB: CDE: F. Wherefore, if four, &c. Q. E. D.