angle ABC is the one half of the parallelogram EBCA; fince the diameter A B cuts into halves, and the triangle DBC the one half of the parallelogram DBCF; because [by prop. 34.] the diameter DC cuts it into halves. But the halves of equal things are themselves equal; therefore the triangle A B C is equal to the triangle D B C.

Therefore triangles conftituted upon the fame base, and between the fame parallels, are the one equal to the other. Which was to be demonftrated.

PROP. XXXVIII. THEOR.

Triangles conftituted upon equal bafes, and between the fame parallels, are equal to one another.

Let the triangles A BC, DEF be conftituted upon the equal bafes B C, E F, and between the fame parallels BF, AD: I fay, the triangle ABC is equal to the triangle

E

F

Then GBC A, D E F H are both parallelograms. B But [by prop. 36.] GB CA is equal to DEF H, for they ftand upon equal bafes BC, EF, between the fame parallels BF, GH. But the triangle ABC is one half of the parallelogram GBCA, becaufe [by prop. 34.] the diameter A B cuts it in halves; and the triangle F ED is the one half of the parallelogram D E F H; for the diameter DF cuts it into halves. But the halves of equal things are themselves equal. Therefore the triangle ABC is equal to the triangle D E F.

Wherefore triangles conftituted upon equal bafes, and between the fame parallels, are equal to one another. Which was to be demonftrated.

PROP

PROP. XXXIX. THEOR.

Equal triangles, conftituted towards the fame parts, upon the fame bafe, are between the fame parallels.

upon

Let the equal triangles ABC, DBC be conftituted the fame bafe B C towards the fame parts: I fay, they are between the same parallels. For draw A D: I fay, this is parallel to B C.

A

E

D

For if not, through the point A [by prop. 31.] draw the right line A E parallel to the right line B C, and draw E C. Then [by prop. 37.] the triangle ABC is equal to the triangle EBC for they are both upon the fame base BC, and between the fame parallels B C, A E. But [by fuppofition] the triangle A B C is equal to the triangle DB C. Therefore alfo the triangle DBC is equal to the triangle E B C, the greater C equal to the lefs; which is impoffible: After the fame man

B

therefore A E is not parallel to в C. ner we demonftrate, that no other line except AD is parallel to B C. Therefore A D is parallel to BC.

Wherefore equal triangles conftituted towards the fame parts, upon the fame bafe, are between the fame parallels. Which was to be demonftrated.

PROP. XL. THEOR.

Equal triangles conftituted towards the fame parts, upon equal bafes, are between the fame parallels.

Let the equal triangles A B C, CDE be conftituted upon equal bafes BC, CE, towards the fame part: I fay, they are also between the fame parallels. For draw A D: I fay, A D is parallel to B E.

For if it be not, through A draw F A parallel to BE, and draw F E.

Then [by prop, 38.] the triangle ABC is equal to the triangle FCE; for they are conftituted upon equal bafes BC, CE, and are between the fame parallels B E, A F, But the triangle ABC is equal to the triangle DCE. And

fo

fo the triangle DCE fhall be equal A
to the triangle FCE, the greater
equal to the leffer; which is im-
poffible. Therefore A F is not pa-
rallel to B E. After the fame man-
ner we demonftrate, that there
is no other line except A D, pa-
rallel to BE. Wherefore AD will
be parallel to B E.

B

C

F

Therefore equal triangles conftituted upon equal bafes, towards the fame parts, are between the fame parallels. Which was to be demonstrated.

If a parallelogram and a triangle have the fame bafe, and are between the fame parallels, the parallelogram will be double to the triangle.

For let the parallelogram A B C D, and the triangle EBC have the fame base BC, and be between the fame parallels BC, AE: I fay, the parallelogram ABCD will be double to the triangle B E C.

For draw A C.

E

Then the triangle ABC [by prop. 37.] is equal to the triangle EBC; for it is conftituted A upon the fame bafe BC, and between the fame parallels B C, A E. But the parallelogram A B C D is double to the triangle ABC; for the diameter A C [by prop. 34.] cuts it into halves: wherefore the parallelogram A B C D will be double to the triangle E B C.

B C

If therefore a parallelogram and a triangle have the fame bafe, and are between the fame parallels, the parallelogram will be double to the triangle, Which was to be demonftrated.

PROP. XLII. PROBL.

To conftitute a parallelogram equal to a given triangle, and having one of its angles equal to a given right-lined angle.

Let the given triangle be A B C, and the given right

lined angle D; it is required to conftitute a parallelogram equal to the triangle A B C, having one of its angles equal to the given right-lined angle D. Bifect B C [by prop. 10.] in E, and draw A E; and with

A

B E

F G

D

the right line E C, at the point E therein [by prop. 23.] make the angle CEF equal to the angle D. And [by prop. 31.] through A, draw A G parallel to Ec; and through c, draw C G parallel to FE: then is FE CG the parallelogram required.

For because BE is equal to E c, the triangle ABE [by prop. 38.] will be equal to the triangle A EC; for they ftand upon equal bafes в E, E C, and are between the fame parallels B C, AG. Therefore the triangle A B C is double to the triangle A E C. But [by prop. 41.] the parallelogram FECG is double to the triangle A EC; for it has the fame bafe, and is between the fame parallels: Therefore [by ax. 6.] the parallelogram FE C G is double to the triangle ABC, and has the angle c E F equal to the given angle D

Therefore the parallelogram F E C G is made equal to the triangle ABC, having the angle c E F equal to the given angle D. Which was to be done.

PROP. XLIII. THEOR.

In every parallelogram, the complements of those parallelograms, which are about a diameter, are equal to one another ".

Let there be a parallelogram A B C D, whofe diameter is A c; and let the parallelograms E H, F G be about the fame; then BK, KD are called the complements of them: I fay, the complement в K is equal to the complement K D.

For because A B C D is a parallelogram, and A c is its diameter, the triangle ABC [by prop. 34.] is equal to the triangle AD C. Again, becaufe EKHA is a parallelogram, whofe diameter is AK, the triangle A EK will be equal to the triangle A HK. For the fame reafon the triangle KFC is equal to the triangle KG C. Therefore fince the triangle

AEK

AEK is equal to the trian- B G

gle AHK, and the triangle KFC to the triangle KG C;

C

the triangle A EK, together with the triangle KG C, is equal to the triangle AHK, together with the triangle KFC. But the whole tri

angle ABC is alfo equal to the whole triangle ADC: Therefore the complement B K remaining, is [by ax. 3.] equal to the complement K D remaining.

Therefore in every parallelogram the complements of thofe parallelograms which are about a diameter, are equal to one another. Which was to be demonftrated.

It may, perhaps, be as well to exprefs this propofition thus: If two right lines be drawn through a point in the diagonal of any parallelogram parallel to the fides, thereby making four leffer parallelograms within the greater, those two of the four leffer parallelograms, which fall without that diagonal, will be equal to one another.

PROP. XLIV. PROBL.

To apply a parallelogram to a given right line equal to a given triangle, and having one angle equal to a given right-lined angle.

Let the given right line be A B, the given triangle c, and the given right-lined angle D: It is required to apply a parallelograrn to the given right line A B, equal to the given triangle c, having one angle equal to D.

Conftitute [by prop. 42.] the parallelogram B E F G

and put B E in a right line with A B. Produce

equal to the triangle c,

F

E

having one angle E BG

equal to the angle D;

Then because

FG to H, and through A draw [by prop. 31.] A H parallel to BG, or E F. Join H B.

That is, to make fuch a parallelogram, that a given right line fhall be one of its fides; one of its angles shall be equal to a given right-lined angle, and the farallelogram shall be equal to a given triangle.

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