to L; ex æquali, ás A to EG, so is H to L: And the figures A, EG, are similar, and M is a mean proportional between H and L; therefore, as was shown in the prece-' ding proposition, CD is to EF as H to M.

PROP. LX.

If a rectilineal figure be given in species and magnitude, the sides of it shall be given in magnitude.

Let the rectilineal figure A be given in species and magnitude, its sides are given in magnitude.

55.

Take a straight line BC given in position and magnitude and upon BC describe the figure D similar, and similarly⚫ 18. 6. placed, to the figure A,

and let EF be the side
of the figure A homo- G
ogous to BC the side

E

of D; therefore the figure D is given in species. And because up

on the given straight

M

line BC the figure D

given in species is described, D is given in magnitude, ' 56 Dat. and the figure A is given in magnitude, therefore the ratio of A to D is given: And the figure A is similar to D: therefore the ratio of the side EF to the homologous side BC is given; and BC is given, wherefored EF is given: $58 Dat. And the ratio of EF to EG is given, therefore EG is 2 Dat. given. And, in the same manner, each of the other sides of the figure A can be shown to be given.

PROBLEM.

To describe a rectilineal figure A similar to a given figure D, and equal to another given figure H. It is Prop. 25, B. 6. Elements.

3 Def.

Because each of the figures D, H, is given, their ratio is given, which may be found by making upon the given Cor. 45. 1. straight line BC the parallelogram BK equal to D, and upon its side CK making the parallelogram KL equal to H in the angle KCL equal to the angle MBC; therefore the ratio of D to H, that is, of BK to KL, is the same with the ratio of BC to CL: And because the figures D, A, are similar, and that the ratio of D to A, or H, is the same with the ratio of BC to CL; by the 58th dat. the ratio of the homologous sides BC, EF, is the same with the ratio of BC to the mean proportional between BC and CL. Find EF

the mean proportional; then EF is the side of the figure to be described, homologous to BC the side of D, and the figure itself can be described by the 18th Prop. B. 6, which by the construction, is similar to D; and because D is to 2 Cor. A, ass BC to CL, that is, as the figure BK to KL; and 20. 6. that D is equal to BK, therefore Ab is equal to KL, that is, to H.

h'14. 5.

See N. IF a parallelogram given in magnitude has one of its sides and one of its angles given in magnitude, the other side also is given.

Let the parallelogram ABDC given in magnitude, have the side AB and the angle BAC given in magnitude, the other side AC is given.

Take a straight line EF given in position and magnitude; and because the parallelogram AD is given in magnitude, a rectilineal figure equal to it can

1 Def. be found. And a parallelogram Cor. 45. 1. equal to this figure can be appliedb

E

A

C

F

B

to the given straight line EF in an angle equal to the given angle BAC. Let this be the parallelogram EFHG, having the angle FEG equal to the angle BAC. And because the parallelograms AD, EH, are equal, and have the angles at A and E equal; the sides about them are reci14. 6. procally proportional; therefore as AB to EF, so is EG to AC: And AB, EF, EG, are given, therefore also AC is 12. 6. givend. Whence the way of finding AC is manifest.

G

H

See N. IF a parallelogram has a given angle, the rectangle, contained by the sides about that angle has a given ratio to the parallelogram.

Let the parallelogram ABCD have the given angle ABC, the rectangle AB, BC, has a given ratio to the parallelogram AC.

From the point A draw AE perpendicular to BC; because the angle ABC is B given, as also the angle AEB, the triangle 43 Dat. ABE is given a in species; therefore the

ratio of BA to AE is given. But as BA to 1. 6. AE, so is the rectangle AB, BC, to the rectangle AE, BC, therefore the ratio of

E

F

GK

H

D

с

the rectangle AB, BC, to AE, BC, that is, to the paral- © 35. 1. lelogram AC, is given.

And it is evident how the ratio of the rectangle to the parallelogram may be found, by making the angle FGH equal to the given angle ABC, and drawing, from any. point F in one of its sides, FK perpendicular to the other GH: for GF is to FK, as BA to AE, that is, as the rectangle AB, BC, to the parallelogram AC.

COR. And if a triangle ABC has a given angle ABC, 66. the rectangle AB, BC, contained by the sides about that angle, shall have a given ratio to the triangle ABC.

Complete the parallelogram ABCD; therefore, by this proposition, the rectrangle AB, BC, has a given ratio to the parallelogram AC; and AC has a given ratio to its half the triangled ABC: therefore the rectangle AB, BC, 1 41. 4. has a given ratio to the triangle ABC.

And the ratio of the rectangle to the triangle is found thus: Make the triangle FGK, as was shown in the proposition: the ratio of GF to the half of the perpendicular FK, is the same with the ratio of the rectangle AB, BC, to the triangle ABC. Because, as was shown, GF is to FK, as AB, BC, to the parallelogram AC; and FK is to its half, as AC is to its half, which is the triangle ABC: therefore, ex æquali, GF is to the half of FK, as AB, BC, rectangle is to the triangle ABC.

PROP. LXIII.

Ir two parallelograms be equiangular, as the side of the first to a side of the second, so is the other side of the second to the straight line to which the other side of the first has the same ratio which the first parallelogram has to the second. And consequently, if the ratio of the first parallelogram to the second be given, the ratio of the other side of the first to that straight line is given; and if the ratio of the other side of the first to that straight line be given, the ratio of the first parallelogram to the second is given.

Let AC, DF, be two equiangular parallelograms; as BC,: a side of the first, is to EF, a side of the second, so is DE, the other side of the second, to the straight line to which AB, the other side of the first, has the same ratio which AC has to DF.

d

e 9 Dat

56.

B

H

Produce the straight line AB, and make as BC to EF so DE to BG, and complete the parallelogram BGHC; therefore, because BC or GH, is to EF, as DE to BG, the sides about the equal angles BGH, DEF, are reciprocally proportional; G * 14, 6. wherefore the parallelogram BH is equal to DF; and AB is to BG, as the parallelogram AC is to BH, that is, to DF; as therefore BC is to EF, so is F DE to BG, which is the straight line to which AB has the same ratio that AC has to DF.

F

And if the ratio of the parallelogram AC to DF be given, then the ratio of the straight line AB to BG is given; and if the ratio of AB to the straight line BG be given, the ratio of the parallelogram AC to DF is given.

ь

See N. Ir two parallelograms have unequal but given angles, and if as a side of the first to a side of the second, so the other side of the second be made to a certain straight line; if the ratio of the first parallelogram to the second be given, the ratio of the other side of the first to that straight line shall be given. And if the ratio of the other side of the first to that straight line be given, the ratio of the first parallelogram to the second shall be given.

Let ABCD, EFGH, be two parallelograms which have the unequal but given angles ABC, EFG; and as BC to FG, so make EF to the straight line M. If the ratio of the parallelogram AC to EG be given, the ratio of AB to M is given.

At the point B of the straight line BC make the angle CBK equal to the angle EFG, and complete the parallelogram KBCL. And because the ratio of AC to EG is given, 35. 1. and that AC is equal to the parallelogram KC, therefore the ratio of KC to EG is given; and as KC, EG, are equi63 Dat. angular, therefore as BC to FG, so is EF to the straight line to which KB has a given ratio, viz. the same which the parallelogram KC has to EG; but as BC to FG, so is EF to the straight line M; therefore KB has a given ratio to M;

Κ. Α.

L Ꭰ

с

43 Dat.

and the ratio of AB to BK is given, because the triangle ABK
is given in species; therefore the ratio of AB to M is givend,
And if the ratio of AB to M be given, the ratio of the
parallelogram AC to EG is given; for since the ratio of
KB to BA is given, as also the ratio
of AB to M, the ratio of KB to M
is given; and because the parallel-
ograms KC, EG are equiangular, as
BC to FG, so is b EF to the straight
line to which KB has the same ratio
which the parallelogram KC has to
EG: But as BC to FG, so is EF to
M; therefore KB is to M, as the pa-
rallelogram KC is to EG; and the ratio of KB to M is
given, therefore the ratio of the parallelogram KC, that is, 75.
of AC, to EG, is given.

B

9 Dat.

b 63 Dat.

E

H

M

F

G

COR. And if two triangles ABC, EFG, have two equal angles, or two unequal but given angles ABC, EFG, and if as BC a side of the first to FG a side of the second, so the other side of the second EF be made to a straight line M; if the ratio of the triangles be given, the ratio of the other side of the first to the straight line M is given.

e

Complete the parallelograms ABCD, EFGH; and because the ratio of the triangle ABC to the triangle EFG is given, the ratio of the parallelogram AC to EG is given, 15, 5, because the parallelograms are doublef of the triangles; and £41, 1. because BC is to FG, as EF to M, the ratio of AB to M is given by the 63d dat. if the angles ABC, EFG, are equal; but if they be unequal but given angles, the ratio of AB to M is given by this proposition.

And if the ratio of AB to M be given, the ratio of the parallelogram AC to EG is given by the same propositions; and therefore the ratio of the triangle ABC to EFG is given.

PROP. LXV.

IF two equiangular parallelograms have a given. ratio to one another, and if one side has to one side a given ratio; the other side shall also have to the other side a given ratio..

Let the two equiangular parallelograms AB, CD, have a given ratio to one another, and let the side EB have a given, ratio to the side FD; the other side AE has also a given ratio to the other side CF.

68.