the ratio of the parallelogram AC to EG is given, and that AC is equal (35. 1.) to BL; therefore the ratio of BL to EG is given and because BL is equiangular to EG, and, by the hypothesis, the ratio of BC to FG is given; therefore (65. dat.) the ratio of KB to EF is given, and the ratio of KB to BA is given; the ratio therefore (9. dat.) of AB to EF is given.

E

F

A K

B

D L

C

M

H

N

G

P

NO

The ratio of AB to EF may be found thus: take the straight line MN given in position and magnitude; and make the angle NMO equal to the given angle BAK, and the angle MNO equal to the given angle EFG or AKB: and because the parallelogram BL is equiangular to FG, and has a given ratio to it, and that the ratio of BC to FG is given; find by the 65th dat. the ratio of KB to EF; and make the ratio of NO to OP the same with it: then the ratio of AB to EF is the same with the ratio of MO to OP: for since the triangle ABK is equiangular to MON, as AB to BK, so is MO to ON: and as KB to EF, so is NO to OP; therefore, ex æquali, as AB to EF, so is MO to OP.

IF the sides of two equiangular parallelograms have given ratios to one another; the parallelograms shall have a given ratio to one another.*

Let ABCD, EFGH be two equiangular parallelograms, and let the ratio of AB to EF, as also the ratio of BC to FG, be given; the ratio of the parallelogram AC to EG is given.

Take a straight line K given in magnitude, and because the ratio of AB to EF is given, make A

DE H

B

C

K

L

F

G

M

the ratio of K to L the same with it; therefore L is given (2. dat.): and because the ratio of BC to FG is given, make the ratio of L to M the same: therefore M is given (2. dat.): and K is given, wherefore (1. dat.) the ratio of K to M is given: but the parallelogram AC is to the parallelogram EG, as the straight line K to the

* See Note.

straight line M, as is demonstrated in the 23d prop. of B. 6. Elem. therefore the ratio of AC to EG is given.

From this it is plain how the ratio of two equiangular parallelograms may be found when the ratios of their sides are given.

If the sides of two parallelograms which have unequal, but given angles, have given ratios to one another; the parallelograms shall have a given ratio to one another.*

Let two parallelograms ABCD, EFGH, which have the given unequal angles ABC, EFG, have the ratios of their sides, viz. of AB to EF,and of BC to FG, given; the ratio of the parallelogram AC to EG is given.

KA

LDE H

At the point B of the straight line BC make the angle CBK equal to the given angle EFG, and complete the parallelogram KBCL; and because each of the angles BAK, BKA is given, the triangle ABK is given (43. dat.) in species: therefore the ratio of AB to BK is given; and the ratio of AB to EF is given, wherefore (9. dat.) the ratio of BK to EF is given: and the ratio of BC to FG is given; and the angle KBC is equal to the angle EFG; therefore (67. dat.) the ratio of the parallelogram KC to EG is given but KC is equal (35. 1.) to AC; therefore the ratio of AC to EG is given.

B

C

M

N

O P Q

F G

The ratio of the parallelogram AC to EG may be found thus: take the straight line MN given in position and magnitude, and make the angle MNO equal to the given angle KAB, and the angle NMO equal to the given angle AKB, or FEH: and because the ratio of AB to EF is given, make the ratio of NO to P the same; also make the ratio of P to Q the same with the given ratio of BC to FG, the parallelogram AG is to EG, as MO to Q.

Because the angle KAB is equal to the angle MNO, and the angle AKB equal to the angle NMO; the triangle AKB is equiangular to NMO therefore as KB to BA, so is MO to ON; and as BA to EF, so is NO to P; wherefore, ex æquali, as KB to EF, so is MO to P: and BC is to FG, as P to Q, and the

* See Note.

parallelograms KC, EG are equiangular; therefore, as was shown in prop. 67, the parallelogram KC, that is, AC, is to EG, as MO to Q.

COR. 1. If two triangles, ABC, DEF have two equal angles, or two unequal, but given angles, ABC, DEF, and if the ratios of the sides about these angles, viz. the ratios of AB to DE, and of BC to EF be given; the triangles shall have a given ratio to one another.

B

A

GD H

C E

F

Complete the parallelograms BG, EH: the ratio of BG to EH is given (67. or 68. dat.); and therefore the triangles which are the halves (34. 1.) of them have a given (15. 5. 72.) ratio to one another.

K A

L D

空空

COR. 2. If the bases BC, EF of two triangles ABC, DEF have a given ratio to one another, and if also the straight lines AG, DH which are drawn to the bases from the opposite angles, either in equal angles, or unequal, but given angles, AGC, DHF have a given ratio to one another; the triangles shall have a given ratio to one another. Draw BK, EL parallel to AG, DH, and complete the parallelograms KC, LF. And because the angles AGC, DHF, or their equals, the angles KBC, LEF are either equal, or unequal, but given; and that the ratio of AG to DH, that is, of KB to LE, is given, as also the ratio of BC to EF; therefore (67. or 68. dat.) the ratio of the parallelogram KC to LF is given; wherefore also the ratio of the triangle ABC to DEF is given (41. 1. 15. 5.).

B G C

E H F

If a parallelogram which has a given angle be applied to one side of a rectilineal figure given in species; if the figure have a given ratio to the parallelogram, the parallelogram is given in species.

Let ABCD be a rectilineal figure given in species, and to one side of it AB, let the parallelogram ABEF, having the given angle ABE, be applied; if the figure ABCD have a given ratio to the parallelogram BF, the parallelogram BF is given in species.

Through the point A draw AG parallel, to BC, and through the point C draw CG parallel to AB, and produce GA, CB to the points H, K: because the angle ABC is given (3. def.), and the ratio of AB to BC is given, the figure ABCD being given in species; therefore the parallelogram BG is given (3. def.) in species. And because upon the same straight line AB the two rectilineal figures BD, BG given in species are described, the ratio of BD to BG is given (53. dat.); and, by hypothesis, the ratio of BD to the parallelogram BF is given; wherefore (9. dat.) the ratio of BF, that is, (35. 1.) of the parallelogram BH, to BG is given, and therefore (1. 6.) the ratio of the straight line KB to BC is given; and the ratio of BC to BA is given, wherefore the ratio of KB to BA is given (9. dat.): and because the angle ABC is given, the adjacent angle ABK is given; and the angle ABE is given, therefore the remaining angle KBE is given. The angle EKB is also given, because it is equal to the angle ABK; therefore the triangle BKE is given in species, and consequently the ratio of EB to BK is given; and the ratio of KB to BA is given, wherefore (9. dat.) the ratio of EB to BA is given; and the angle ABE is given, therefore the parallelogram BF is given in species.

L

D

N

G

C

мо

B

A

H F K E

R

A parallelogram similar to BF may be found thus: take a straight line LM given in position and magnitude; and because the angles ABK, ABE are given, make the angle NLM equal to ABK, and the angle NLO equal to ABE. And because the ratio of BF to BD is given, make the ratio of LM to P the same with it; and because the ratio of the figure BD to BG is given, find this ratio by the 53d dat. and make the ratio of P to Q the same. Also, because the ratio of CB to BA is given, make the ratio of Q to R the same; and take LN equal to R; through the point M draw OM parallel to LN, and complete the parallelogram NLOS; then this is similar to the parallelogram BF.

Because the angle ABK is equal to NLM, and the angle ABE to NLO, the angle KBE is equal to MLO; and the angles BKE, LMO are equal, because the angle ABK is equal to NLM; therefore the triangles BKE, LMO are equiangular to one another; wherefore as BE to BK, so is LO to LM; and because as the figure BF to BD, so is the straight line LM to P; and as BD to BG, so is P to Q; ex æquali, as BF, that is (35.

1.) BH to BG, so is LM to Q: but BH is to (1. 6.) BG, as KB to BC: as therefore KB to BC, so is LM to Q; and because BE is to BK, as LO to LM; and as BK to BC, so is LM to Q and as BC to BA, so Q was made to R; therefore, ex æquali, as BE to BA, so is LO to R, that is to LN; and the angles ABE, NLO are equal; therefore the parallelogram BF is similar to LS.

Ir two straight lines have a given ratio to one another, and upon one of them be described a rectilineal figure given in species, and upon the other a parallelogram having a given angle; if the figure have a given ratio to the parallelogram, the parallelogram is given in species.*

Let the two straight lines AB, CD have a given ratio to one another, and upon AB let the figure AEB given in species be described, and upon CD the parallelogram DF having the given angle FCD; if the ratio of AEB to DF be given, the parallelogram DF is given in species.

E

F

B

D

Upon the straight line AB, conceive the parallelogram AG to be described similar, and similarly placed to FD; and because the ratio of AB to CD is given, and upon them are described the similar rectilineal figures AG, FD; the ratio of AG to FD is given (54. dat.); and the ratio of FD to AEB A is given; therefore (9. dat.) the ratio of AEB to AG is given; and the angle ABG is given, because it is equal to the angle FCD: because therefore the parallelogram AG which has a given angle ABG is applied to a side AB of the figure AEB given in species, and the ratio of AEB to AG is given, the parallelogram AG is given (69. dat.) in species; but FD is similar to AG; therefore FD is given in species.

M

N

HK L

A parallelogram similar to FD may be found thus: take a straight line H given in magnitude; and because the ratio of the figure AEB to FD is given, make the ratio of H to K the same with it also, because the ratio of the straight line CD to AB is given, find by the 54th dat. the ratio which the figure FD described upon CD has to the figure AG described upon AB similar to FD; and make the ratio of K to L the same with this ratio;

• See Note.