equal to the inward and opposite angle AEC: but the angle ACE has been proved equal to the angle BAD; therefore also ACE is equal to the angle AEC, and consequently the side AE is equal to the side (6. 1.) AC; and because AD is drawn parallel to one of the sides of the triangle BCE, viz. to EC, BD is to DC, as BA to AE (3.5.): but AE is equal to AC; therefore, as BD to DC, so is BA to AC (13.4.). Let now BD be to DC, as BA to AC, and join AD; the angle BAC is divided into two equal angles by the straight line AD.

The same construction being made; because, as BD to DC, so is BA to AC: and as BD to DC, so is BA to AE (3. 5.), because AD is parallel to EC; therefore BA is to AC, as BA to AE (13. 4.) : consequently AC is equal to AE (14. 4.), and the angle AEC is therefore equal to the angle ACE (5. 1.): but the angle AEC is equal to the corresponding angle BAD: and the angle ACE is equal to the corresponding angle CAD (3.2.): wherefore also the angle BAD is equal to the angle CAD: therefore the angle BAC is cut into two equal angles by the straight line AD. Therefore, if the angle, &c. Q. E. D.

PROP. V. THEOR.

The sides about the equal angles of equiangular triangles are proportionals; and those which are opposite to the equal angles are homologous sides, that is, are the antecedents or consequents of the ratios.

Let ABC, DCE be equiangular triangles, having the angle ABC equal to the angle DCE, and the angle ACB to the angle DEC, and consequently (5. 2.) the angle BAC equal to the angle CDE. The sides about the equal angles of the triangles ABC, DCE are proportionals; and those are the homologous sides which are opposite to the equal angles.

Let the triangle DCE be placed so that its side CE may be contiguous to BC, and in the same straight line with it; and because the angles ABC, ACB are together less than two right angles (5. 2.); ABC and DEC, which is equal to

F

A

ACB, are also less than two right angles: wherefore BA, ED produced shall meet (13.2.); let them be produced and meet in the point F, and because the angle ABC is equal to the angle DCE, BF is parallel (1. Cor. 2. 2.) to CD. Again, because the angle ACB is equal to the angle DEC, AC is parallel to FE (1. Cor. 2. 2.); therefore FACD is a parallelogram; and consequently AF is equal to CD, and AC to FD (9. 2.):

and because AC is parallel to FE, one of the sides of the triangle FBE, BA is to AF, as BC to CE (3.5.): but AF is equal to CD, therefore (13. 4.) as BA to CD, so is BC to CE; and alternately, as AB to BC, so DC to CE. Again, because CD is parallel to BF, as BC to CE, so is FD to DE (3.5.); but FD is equal to AC; therefore as BC to CE, so is AC to DE; and alternately, as BC to CA, so CE to ED: therefore because it has been proved that AB is to BC, as DC to CE; and as BC to CA, so CE to ED, ex æquali (16. 4.), ВА is to AC, as CD to DE. Therefore the sides, &c. Q. E. D.

PROP. VI. THEOR.

If the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular, and have their equal angles opposite to the homologous sides.

Let the triangles ABC, DEF have their sides proportionals, so that AB is to BC, as DE to EF; and BC to CA, as EF to FD; and consequently, ex æquali, BA to AC, as ED to DF; the triangle ABC is equiangular to the triangle DEF, and their equal angles are opposite to the homologous sides, viz. the angle ABC equal to the angle DEF, and BCA to EFD, and also BAC to EDF.

At the points E, F, in the straight line EF, make (14. 1.) the angle FEG equal to the angle ABC, and the angle EFG equal to BCA;

wherefore the remaining angle

BAC is equal to the remaining angle EGF (5. 2.), and the triangle ABC is therefore equiangular to the triangle GEF; and consequently they have their sides opposite to the equal angles proportionals (5.5.). Wherefore, as AB to BC, so is GE to EF; but as AB to BC, so is DE to EF; therefore as DE to EF,

1

so (15. 4.) GE to EF: therefore DE and GE have the same ratio to EF, and consequently are equal (14.4.): for the same reason, DF is equal to FG: and because in the triangles DEF, GEF, DE is equal to EG, and EF common, the two sides DE, EF are equal to the two GE, EF, and the base DF is equal to the base GF: therefore the angle DEF is equal (7. 1.) to the angle GEF, and the other angles to the other angles which are subtended by the equal sides (4.1.): wherefore the angle DFE is equal to the angle GFE, and EDF to EGF: and because the angle DEF is equal to the angle GEF, and GEF to the angle ABC; therefore the angle ABC is equal to the angle DEF: for the same reason the angle ACB is equal to the angle DFE, and the angle at A to the angle at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the sides, &c. Q, E. D.

1

PROP. VII. THEOR.

If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.

Let the triangles ABC, DEF have the angle BAC in the one equal to the angle EDF in the other, and the sides about those angles proportionals; that is, BA to AC, as ED to DF; the triangles ABC, DEF are equiangular, and have the angle ABC equal to the angle DEF, and ACB to DFE.

A

AN

CEF

At the points D, F, in the straight line DF, make (14. 1.) the angle FDG equal to either of the angles BAC, EDF; and the angle DFG equal to the angle ACB; wherefore the remaining angle at B is equal to the remaining one at G (5. 2.), and consequently the triangle ABC is equiangular to the triangle DGF; and therefore as BA to AC, so is (5. 5.) GD to DF; but, by the hypo- B thesis, as BA to AC, so is ED to DF; as therefore ED to DF, so is (15. 4.) GD to DF; wherefore ED is equal (14.4.) to DG; and DF is common to the two triangles EDF, GDF; therefore the two sides ED, DF are equal to the two sides GD, DF: and the angle EDF is equal to the angle GDF; wherefore the base EF is equal to the base FG (4. 1.), and the triangle EDF to the triangle GDF, and the remaining angles to the remaining angles, each to each, which are subtended by the equal sides; 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 to the angle ACB; therefore the angle ACB is equal to the angle DFE: and the angle BAC is equal to the angle EDF (Hyp.); wherefore also the remaining angle at B is equal to the remaining angle at E. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q. E. D.

PROP. VIII. THEOR.

In a right angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar 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 similar to the whole triangle ABC, and to one another.

Because the angle BAC is equal to the angle ADB, each of them being a right angle, and that the angle at Bis common to the two triangles ABC, ABD; the re

maining angle ACB is equal to

the remaining angle BAD (5.2.): therefore the triangle ABC is equiangular to the triangle ABD, and the sides about their equal angles are proportionals (5. 5.); wherefore the triangles are similar (5 Def. 5.): in the like manner it may be demonstrated, that the triangle ADC is equiangular and similar to the triangle ABC: and the triangles ABD, ADC, being both equiangular and similar to ABC, are equiangular and similar to each other. Therefore, in a right angled, &c. Q. E. D.

B

DC

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: because in the triangles BDA, ADC, BD is to DA as DA to DC (5. 5.); and in the triangles ABC, DBA, BC is to BA, as BA to BD (5. 5.); and in the triangles ABC, ACD, BC is to CA as CA to CD (5. 5.).

PROP. IX. PROB.

From a given straight line to cut off any part required.

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 takeAC the same multiple

of AD, that AB is of the part which is to be

cut off from it: join BC, and draw DE paral

lel to it: then AE is the part required to be cut off.

A

Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, as CD is to DA, so is (3. 5.) BE to EA; and, by composition (7. 4.) CA is to AD as BA to AE: but CA is a multiple of AD; therefore (Def. 18.

ED

4.) BA is the same multiple of AE: whatever
part therefore AD is of AC, AE is the same
part of AB: wherefore, from the straight line
AB the part required is cut off. Which was to be done.

P

C

PROP. X. PROB.

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 in 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 (4. 2.) DF, EG parallels to it;

and through D draw DHK parallel to AB: therefore each of the figures FH, HB, is a parallelogram; wherefore DH is equal (9. 2.) to FG, and HK to GB: and because

A

HE is parallel to KC, one of the sides of F

G

B

D

H

E

K

the triangle DKC, as CE to ED, so is (3. 5.) KH to HD; but KH is equal to BG, and HD to GF; therefore as CE to ED, so is BG to GF: again, because FD is parallel to EG, one of the sides of the triangle AGE, as ED to DA, so is GF to FA; but it has been proved that CE is to ED, as BG to GF; as therefore CE to ED, so is BG to GF; and as ED to DA, so GF to FA: 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 required to find a third proportional to AB, AC.

A

Produce AB, AC to the points D, E ; and make BD equal to AC, and having joined BC, through D draw DE paral- Bв ✓ lel to it (4. 2.).

C

Because BC is parallel to DE, a side
of the triangle ADE, AB is (3. 5.) to
BD, as AC to CE: but BD is equal to
AC; as therefore AB to AC, so is AC to
CE. Wherefore to the two given straight D

E

lines AB, AC, a third proportional CE is