B

Now the

DH

KLM

GL

K F

same as the ratios of the sides, viz. of BC to CG and DC to CE: But the ratio of K to M (5. Def. A) is that which is said to be compounded of the ratios of K to L and L to M; therefore also the ratio of K to M is the ratio compounded of the ratios of the sides: parallelogram AC is to the parallelogram A ICH as BC to CG; but BC is to CG as K to L; therefore also the parallelogram AC is to the parallelogram CH as K to L: Again, the parallelogram CH is to the parallelogram CF as DC to CE; but DC is to CE as L to M; therefore also the parallelogram CH is to the parallelogram CF as L to M: Since, then, it has been proved that the parallelogram AC is the parallelogram CH as K to L, and the parallelogram CH to the parallelogram CF as L to M, therefore, ex æquali, the parallelogram AC is to the parallelogram CF as K to M: But K is to M in the ratio which is compounded of the ratios of the sides; therefore also the parallelogram AC is to the parallelogram CF in the ratio which is compounded of the ratios of the sides.

Wherefore, Equiangular parallelograms &c. Q.E.D.

PROP. XXIV. THEOR.

The parallelograms, which are about the diameter of any parallelogram, are similar to the whole and to one another.

Let ABCD be a parallelogram, of which the diameter is AC, and EG, HK the parallelograms about the diameter: the parallelograms EG, HK shall be similar to the whole parallelogram ABCD, and to one another. For, because DC, GF are parallels, the angle ADC is equal to the angle AGF: For the like reason, because

BC, EF are parallels, the angle ABC is equal to the angle AEF: And each of the angles BCD, EFG is equal to the opposite angle BAD, and therefore to one another: Therefore the parallelograms ABCD, AEFG are equiangular: And because the angle ABC is equal to the angle AEF, and the angle BAC common to the two triangles BAC, EAF, therefore they are equiangular and similar to one another, and AB is to BC as AE to EF: But the opposite sides of parallelograms are equal to one another; therefore AB is to AD as AE to AG, and CD to CB as FG FE, and CD to DA as FG to GA: Therefore the parallelograms ABCD, AEFG have the sides about their equal angles proportionals, and are therefore similar to one another: For the same reason the parallelograms ABCD, FHCK are similar to one another: Therefore each of the parallelograms EG, HK is similar to BD, and therefore (6. 21) the parallelogram EG is similar to the parallelogram KH.

E

B

G

F

H

D K

Wherefore, The parallelograms &c. Q.E.D.

PROP. XXV.

PROB.

C

To describe a rectilineal figure which shall be similar to one given rectilineal figure and equal to another.

Let ABC be the given rectilineal figure to which the figure to be described is to be similar, and D that to which it is to be equal: it is required to describe a rectilineal figure, similar to ABC and equal to D.

Upon BC describe the parallelogram BE equal to the figure ABC, and upon CE describe the parallelogram CM equal to D, and having the angle FCE equal to the angle CBL; therefore BC and CF will be in the same straight line, as also LE and EM: between BC and CF

find a mean proportional GH (6. 13), and upon GH describe the rectilineal figure KGH, similar and similarly situated to the figure ABC (6. 18): Then, because BC is to GH as GH to CF, and that, if three straight lines be proportionals, as the first is to the third, so is any figure upon the first to a similar and similarly described figure upon the second, therefore, as BC to CF, so is the figure ABC to the figure KGH: But, as BC to CF, so is the parallelogram BE to the parallelogram EF; therefore the figure ABC is to the figure KGH as the parallelogram BE to the parallelogram EF: And the figure ABC is equal to the parallelogram BE; therefore (5. 14) the figure KGH is equal to the parallelogram EF : But the parallelogram EF is equal to the figure D; therefore also the figure KGH is equal to the figure D, and it is similar to ABC: Therefore the rectilineal figure KGH has been described, similar to the figure ABC and equal to D.

D

K

ARA

B

L

Q. E. F.

PROP. XXVI. THEOR.

C F

B M

G H

If two similar parallelograms have a common angle and be similarly situated, they are about the same diameter. Let the parallelograms ABCD, AEFG be similar and similarly situated, and have the common angle BAD: ABCD and AEFG shall be about the same diameter. For, if not, let, if possible, the parallelogram BD have its diameter AHC in a different straight line from AF, the diameter of the parallelogram EG; let GF meet AHC in H, F and through H draw HK parallel to AD or BC: Then the parallelograms ABCD,

K

A

D

H

B

F

AKHG, being about the same diameter, are similar to one another (6. 24); therefore DA is to AB as GA to AK: But, because ABCD and AEFG are similar parallelograms, DA is to AB as GA to AE: Therefore GA is to AK as GA to AE, and, consequently (5. 9), AK is equal to AE, the less to the greater-which is impossible: Therefore the parallelograms ABCD, AEFG cannot but have their diameters in the same straight line, that is, they are about the same diameter. Wherefore, If two similar &c. Q.E. D.

PROP. XXVII:-XXIX.

PROP. XXX. PROB.

To cut a given straight line in extreme and mean ratio.

Let AB be the given straight line: it is required to cut it in extreme and mean ratio.

ACB

Divide AB in the point C, so that the rectangle contained by AB, BC, may be equal to the square of AC (2. 11): Then, because the rectangle AB, BC is equal to the square of AC, therefore (6. 17) AB to AC as AC to CB: therefore (6. Def. 3) AB

is cut in extreme and mean ratio in C.

Q. E. F.

In any right-angled triangle, any rectilineal figure described upon the side subtending the right angle is equal to the similar and similarly situated figures upon the sides containing the right angle.

Let ABC be a right-angled triangle, having the right angle BAC, and upon AB, AC, BC let any similar and

similarly situated figures be described: the figure upon BC shall be equal to the similar and similarly situated figures upon BA, AC.

Draw the perpendicular AD: Then, because in the right-angled triangle ABC, AD is drawn, from the right angle at A, perpendicular to the base BC, therefore (6. 8) the triangles ADB, ADC are similar to the whole triangle ABC, and to one another: And because the triangle ABC is similar to the triangle ABD, therefore CB is to BA as BA to BD: But when three straight lines are proportionals, as the first is to the third, so is any figure upon the first to the similar and similarly described figure upon the second (6. 20. Cor.); therefore as CB to BD, so is the figure upon CB to the similar and similarly situated figure upon BA, and inversely, as BD to BC, so is the figure upon BA to the figure upon BC: In like manner, as CD to CB, so is the figure upon CA to the figure upon CB: Therefore (5. 24) as BD and CD together to BC, so are the

D

figures upon BA, AC together to the figure upon BC: But BD and CD together are equal to BC; therefore (5. A.) the figure upon BC is equal to the similar and similarly situated figures on BA, AC.

Wherefore, In any right-angled triangle &c.

Q. E. D.

PROP. XXXII. THEOR.

If two triangles, which have two sides of the one proportional to two sides of the other, be joined at one angle, so

as to have their homologous sides parallel to one another,

the remaining sides shall be in a straight line.

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