Sidebilder
PDF
ePub

PROPOSITION XXXV. THEOREM. Parallelograms upon the same base, and between the same parallels, are equal to one another.

Let the parallelograms ABCD, EBCF be upon the same base BC, Y and between the same parallels AF, BC.

Then the parallelogram ABCD shall be equal to the parallelogram EBCF

A D F A DE F A E DI

в с

B C

. в с - If the sides AD, DF of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated in the same point D; then it is plain that each of the parallelograms is double of the triangle

BDC; (1. 34.) and therefore the parallelogram ABCD is equal to the parallelogram DBCF. (ax. 6.) wa But if the sides AD, EF, opposite to the base BC, be not terminated in the same point;

Then, because ABCD is a parallelogram,

therefore AD is equal to BC; (1. 34.)
and for a similar reason, EF is equal to BC;
wherefore AD is equal to EF; (ax. 1.)

and DE is common; therefore the whole, or the remainder AE, is equal to the whole, or the remainder DF; (ax. 2 or 3.)

and AB is equal to DC; (1. 24.) hence in the triangles EAB, FDC,

because FD is equal to EA, and DC to AB, and the exterior angle FDC is equal to the interior and opposite angle EAB ; (1. 29.)

therefore the base FC is equal to the base EB, (1. 4.) and the triangle FDC is equal to the triangle EAB. From the trapezium ABCF take the triangle FDC, and from the same trapezium take the triangle EAB,

and the remainders are equal, (ax. 3.) therefore the parallelogram ABCD is equal to the parallelogram EBCF.

Therefore, parallelograms upon the same, &c. Q.E.D.

PROPOSITION XXXVI. THEOREM. Parallelograms upon equal bases and between the same parallels, are equal to one another.

Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and between the same parallels AH, BG.

Then the parallelogram ABCD shall be equal to the parallelogram EFGH.

[blocks in formation]

с F

Join BE, CH. Then because BC is equal to FG, (hyp.) and FG to EH, (1. 34.)

therefore BC is equal to EH; (ax. 1.) and these lines are parallels, and joined towards the same parts by the

straight lines BE, CH; but straight lines which join the extremities of equal and parallel straight lines towards the same parts, are themselves equal and parallel ; (1. 33.)

therefore BE, CH are both equal and parallel ;

wherefore EBCH is a parallelogram. (def. A.) And because the parallelograms ABCD, EBCH, are upon the

same base BC, and between the same parallels BC, AH; therefore the parallelogram ABCD is equal to the parallelogram EBCH. (I. 35.)

For the same reason, the parallelogram EFGH is equal to the parallelogram EBCH; . therefore the parallelogram ABCD is equal to the parallelogram EFGH. (ax. 1.) Therefore, parallelograms upon equal, &c. Q.E.D.

PROPOSITION XXXVII. THEOREM. Triangles upon the same base and between the same parallels, are equal to one another. Let the triangles ABC, DBC be upon the same base BC,

and between the same parallels AD, BC. Then the triangle ABC shall be equal to the triangle DBC.

E AD F

Produce AD both ways to the points E, F;
through B draw BE parallel to CA, (1. 31.)

and through C draw CF parallel to BD. Then each of the figures EBCA, DBCF is a parallelogram ; and EBCA is equal to DBCF, (1. 35.) because they are upon the

same base BC, and between the same parallels BC, EF. And because the diameter AB bisects the parallelogram EBCA, therefore the triangle ABC is half of the parallelogram EBCA; (1. 34.)

also because the diameter DC bisects the parallelogram DBCF, therefore the triangle DBC is half of the parallelogram DBCF,

but the halves of equal things are equal ; (ax. 7.) therefore the triangle ABC is equal to the triangle DBC. Wherefore, triangles, &c. Q.E.D.

PROPOSITION XXXVIII. THEOREM. Triangles upon equal bases and between the same parallels, are equal to one another.

Let the triangles ABC, DEF be upon equal bases BC, EF, and between the same parallels BF, AD. Then the triangle ABC shall be equal to the triangle DEF.

G A D I

B CE F
Produce AD both ways to the points G, II;
through B draw BG parallel to CA, (1. 31.)

and through F draw FH parallel to ED.
Then each of the figures GBCA, DEFH is a parallelogram;

and they are equal to one another, (1. 36.)
because they are upon equal bases BC, EF,

and between the same parallels BF, GH. And because the diameter AB bisects the parallelogram GBCA, therefore the triangle ABC is the half of the parallelogram GBCA; (1. 34.) . also, because the diameter DF bisects the parallelogram DEFH, therefore the triangle DEF is the half of the parallelogram DEFH;

but the halves of equal things are equal; (ax. 7.) therefore the triangle ABC is equal to the triangle DEF.

Wherefore, triangles upon equal bases, &c. Q. E. D.

PROPOSITION XXXIX. TIIEOREM. Equal triangles upon the same base and upon the same side of it, are between the same parallels.

Let the equal triangles ABC, DBC be upon the same base BC, and upon the same side of it. Then the triangles ABC, DBC shall be between the same parallels.

в
Join AD; then AD shall be parallel to BC.

For if AD be not parallel to BC, if possible, through the point A, draw A E parallel to BC, (1. 31.) meeting BD, or BD produced, in E, and join EC. Then the triangle ABC is equal to the triangle EBC, (1. 37.)

because they are upon the same base BC,

and between the same parallels BC, AE:
but the triangle ABC is equal to the triangle DBC; (hyp.)
therefore the triangle DBC is equal to the triangle EBC,

the greater triangle equal to the less, which is impossible:

therefore AE is not parallel to BC.

In the same manner it can be demonstrated,
that no other line drawn from A but AD is parallel to BC;

AD is therefore parallel to BC.
Wherefore, equal triangles upon, &c. Q. E. D.

PROPOSITION XL. THEOREM. Equal triangles upon equal bases in the same straight line, and towards the same parts, are between the same parallels.

Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight line BF, and towards the same parts. Then they shall be between the same parallels.

D

А

B C E F
Join AD; then AD shall be parallel to BF.

For if AD be not parallel to BF,
if possible, through A draw À G parallel to BF, (1. 31.)

meeting ED, or ED produced in G, and join ĞF.
Then the triangle ABC is equal to the triangle GEF, (1. 38.)

because they are upon equal bases BC, EF,

and between the same parallels BF, AG; but the triangle ABC is equal to the triangle DEF; (hyp.) therefore the triangle DEF is equal to the triangle GEF, (ax. 1.) the greater triangle equal to the less, which is impossible:

therefore AG is not parallel to BF. And in the same manner it can be demonstrated, that there is no other line drawn from A parallel to it but AD;

AD is therefore parallel to BF.
Wherefore, equal triangles upon, &c. Q. E. D.

PROPOSITION XLI. THEOREM. If a parallelogram and a triangle be upon the same base, and between the same parallels ; the parallelogram shall be double of the triangle.

Let the parallelogram ABCD, and the triangle EBC be upon the same base BC, and between the same parallels BC, AE. Then the parallelogram ABCD shall be double of the triangle EBC.

. DE

в

Join AC.
Then the triangle ABC is equal to the triangle EBC, (1. 37.)

because they are upon the same base BC, and between the same

parallels BC, AE. But the parallelogram ABCD is double of the triangle ABC,

because the diameter AC bisects it; (1. 34.) wherefore ABCD is also double of the triangle EBC. Therefore, if a parallelogram and a triangle, &c. Q. E. D.

PROPOSITION XLII. PROBLEM. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Let ABC be the given triangle, and D the given rectilineal angle.

It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D.

A F . G

B E C
Bisect BC in E, (1. 10.) and join AE;

at the point E in the straight line EC, make the angle CEF equal to the angle D; (I. 23.) through C draw CG parallel to EF, and through A draw AFG parallel to BC, (1. 31.) meeting EF in F, and CG in G.

Then the figure CEFG is a parallelogram. (def. A.) And because the triangles ABE, AEC are on the equal bases BE, EC, and between the same parallels BC, AG;

they are therefore equal to one another; (1. 38.)

and the triangle ABC is double of the triangle AEC; but the parallelogram FECG is double of the triangle AEC, (1. 41.) because they are upon the same base EC, and between the same

parallels EC, AG; therefore the parallelogram FECG is equal to the triangle ABC, (ax. 6.)

and it has one of its angles CEF equal to the given angle D. Wherefore, a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D. Q.E.F.

PROPOSITION XLIII. THEOREM. The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.

Let ABCD be a parallelogram, of which the diameter is AC: and EH,GF the parallelograms about AC, that is, through which A C passes: also BK, KD the other parallelograms which make up the whole

figure ABCD, which are therefore called the complements. Then the complement BK shall be equal to the complement KD.

« ForrigeFortsett »