EF is greater than KM;

therefore the straight line EG is greater than KL, and GF than LM: make GX equal to LK, and GO equal to LM, (I. 3.) and complete the parallelogram XGOP: (I. 31.) therefore XO is equal and similar to KM: but KM is similar to EF;

wherefore also XO is similar to EF;

and therefore XO and EF are about the same diameter: (VI. 26.) let GPB be their diameter and complete the scheme. Then, because EF is equal to C and KM together,

and XO a part of the one is equal to KM a part of the other, the remainder, viz. the gnomon ERO, is equal to the remainder C: (ax. 3.)

and because OR is equal to XS, by adding SR to each, (1. 43.) the whole OB is equal to the whole XB: but XB is equal to TE, because the base AE is equal to the base EB; (1. 36.)

wherefore also TE is equal to OB: (ax. 1.)

add XS to each, then the whole TS is equal to the whole, viz. to the gnomon ERO:

but it has been proved that the gnomon ERO is equal to C';

and therefore also TS is equal to C.

Wherefore the parallelogram TS, equal to the given rectilineal figure C, is applied to the given straight line AB, deficient by the parallelogram SR, similar to the given one D, because SR is similar to EF. (VI. 24.) Q. E. F.

PROPOSITION XXIX. PROBLEM.

To a given straight line to apply a parallelogram equal to a given rectilineal figure, exceeding by a parallelogram similar to another given.

Let AB be the given straight line, and C the given rectilineal figure to which the parallelogram to be applied is required to be equal, and D the parallelogram to which the excess of the one to be applied above that upon the given line is required to be similar.

It is required to apply a parallelogram to the given straight line AB which shall be equal to the figure C, exceeding by a parallelogram similar to D.

Divide AB into two equal parts in the point E, (1. 10.) and upon EB describe the parallelogram EL similar and similarly situated to D: (VI. 18.)

and make the parallelogram GH equal to EL and Ctogether, and similar and similarly situated to D: (VI. 25.)

wherefore GH is similar to EL: (VI. 21.)

let KH be the side homologous to FL, and KG to FE:
and because the parallelogram GH is greater than EL,
therefore the side KH is greater than FL,
and KG than FE:

produce FL and FE, and make FLM equal to KH, and FEN to KG,
and complete the parallelogram MN:
MN is therefore equal and similar to GH:
but GH is similar to EL;

wherefore MN is similar to EL;

and consequently EL and MN are about the same diameter: (VI. 26.) draw their diameter FX, and complete the scheme. Therefore, since GH is equal to EL and C together, and that GH is equal to MN;

MN is equal to EL and C:
take away the common part EL;

then the remainder, viz. the gnomon NOL, is equal to C.
And because AE is equal to EB,

the parallelogram AN is equal to the parallelogram NB, (1.36.) that is, to BM: (1.43.)

add NO to each;

therefore the whole, viz. the parallelogram AX, is equal to the gnomon NOL:

but the gnomon NOL is equal to C';

therefore also AX is equal to C.

Wherefore to the straight line AB there is applied the parallelogram AX equal to the given rectilineal figure C, exceeding by the parallelogram PO, which is similar to D, because PO is similar to EL. (vI. 24.) Q.E.F.

PROPOSITION XXX. PROBLEM.

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.

Upon AB describe the square BC, (1.46.)

and to AC apply the parallelogram CD, equal to BC, exceeding by the figure AD similar to BC: (VI. 29.)

then, since BC is a square,

therefore also AD is a square:

and because BC is equal to CD,

by taking the common part CE from each,
the remainder BF is equal to the remainder AD:
and these figures are equiangular,

therefore their sides about the equal angles are reciprocally propor

tional: (VI. 14.)

therefore, as FE to ED, so AE to EB:

but FE is equal to AC, (1. 34) that is, to AB; (def. 30.)
and ED is equal to AE;

therefore as BA to AE, so is AE to EB:
but AB is greater than AE;

wherefore AE is greater than EB: (v. 14.)

therefore the straight line AB is cut in extreme and mean ratio in E. (VI. def. 3.) Q.E.F.

Otherwise,

Let AB be the given straight line.

It is required to cut it in extreme and mean ratio.

Divide AB in the point C, so that the rectangle contained by AB, BC, may be equal to the square on AC. (II. 11.)

Then, because the rectangle AB, BC is equal to the square on AC; as BA to AC, so is AC to CB: (VI. 17.)

therefore AB is cut in extreme and mean ratio in C. (VI. def. 3.)

Q.E.F.

PROPOSITION XXXI. THEOREM.

In right-angled triangles, the rectilineal figure described upon the side opposite to the right angle, is equal to the similar and similarly described figures upon the sides containing the right angle.

Let ABC be a right-angled triangle, having the right angle BAC. The rectilineal figure described upon BC shall be equal to the similar and similarly described figures upon BA, AC.

Draw the perpendicular AD: (1. 12.)

therefore, because in the right-angled triangle ABC,

AD is drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ÅDC are similar to the whole triangle ABC, and to one another: (VI. 8.)

and because the triangle ABC is similar to ADB,

as CB to BA, so is BA to BD: (VI. 4.)

and because these three straight lines are proportionals,

as the first is to the third, so is the figure upon the first to the similar and similarly described figure upon the second: (VI. 20. Cor. 2.) therefore as CB to BD, so is the figure upon CB to the similar and similarly described figure upon BA:

and inversely, as DB to BC, so is the figure upon BA to that upon BC: (V. B.)

for the same reason, as DC to CB, so is the figure upon CA to that upon CB:

therefore as BD and DC together to BC, so are the figures upon BA, AC to that upon BC: (v. 24.)

but BD and DC together are equal to BC; therefore the figure described on BC is equal to the similar and similarly described figures on BA, AC. (v. A.) Wherefore, in right-angled triangles, &c.

Q.E.D.

PROPOSITION XXXII. THEOREM.

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.

Let ABC, DCE be two triangles which have the two sides BA, AC proportional to the two CD, DE,

viz. BA to AC, as CD to DE;

and let AB be parallel to DC, and AC to DE.

A

D

B

Then BC and CE shall be in a straight line.

Because AB is parallel to DC, and the straight line AC meets them, the alternate angles BAC, ACD are equal; (1. 29.) for the same reason, the angle CDE is equal to the angle ACD ; wherefore also BAC is equal to CDE: (ax. 1.)

and because the triangles ABC, DCE have one angle at A equal to one at D, and the sides about these angles proportionals, viz. BA to AC, as CD to DE,

the triangle ABC is equiangular to DCE: (vi. 6.) therefore the angle ABC is equal to the angle DCE: and the angle BAC was proved to be equal to ACD; therefore the whole angle ACE is equal to the two angles ABC, BAC: (ax. 2.)

add to each of these equals the common angle ACB, then the angles ACE, ACB are equal to the angles ÃBC, BAC, ACB : but ABC, BAC, ACB are equal to two right angles: (1. 32.) therefore also the angles ACE, ACB are equal to two right angles: and since at the point C, in the straight line AC, the two straight lines BC, CE, which are on the opposite sides of it, make the adjacent angles ACE, ACB equal to two right angles;

therefore BC and CE are in a straight line. (I. 14.)

Wherefore, if two triangles, &c.

Q.E.D.

PROPOSITION XXXIII. THEOREM.

In equal circles, angles, whether at the centers or circumferences, have the same ratio which the circumferences on which they stand have to one another: so also have the sectors.

Let ABC, DEF be equal circles; and at their centers the angles BGC, EHF, and the angles BAC, EDF, at their circumferences. As the circumference BC to the circumference EF, so shall the angle BGC be to the angle EHF, and the angle BAC to the angle EDF;

and also the sector BGC to the sector EHF.

Take any number of circumferences CK, KL, each equal to BC, and any number whatever FM, MN, each equal to EF: and join GK, GL, HM, HN.

Because the circumferences BC, CK, KL are all equal, the angles BGC, CGK, KGL are also all equal: (III. 27.) therefore what multiple soever the circumference BL is of the circumference BC, the same multiple is the angle BGL of the angle BGC:

for the same reason, whatever multiple the circumference EN is of the circumference EF, the same multiple is the angle EHN of the angle EHF:

and if the circumference BL be equal to the circumference EN, the angle BGL is also equal to the angle EHN; (III. 27.)

and if the circumference BL be greater than EN,

likewise the angle BGL is greater than EHN; and if less, less: therefore, since there are four magnitudes, the two circumferences BC, EF, and the two angles BGC, EHF; and that of the circumference BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the circumference BL, and the angle BGL; and of the circumference EF, and of the angle EHF, any equimultiples whatever, viz. the circumference EN, and the angle EHN:

and since it has been proved, that if the circumference BL be greater than EN;

the angle BGL is greater than EHN;

and if equal, equal; and if less, less;

therefore as the circumference BC to the circumference EF, so is the angle BGC to the angle EHF: (v. def. 5.)

but as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF: (v. 15.)

for each is double of each; (III. 20.)

therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.