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Then because AB is parallel to CD [Hyp.] and

BC meets them, the alternate angles ABC, BCD are equal. [I. 29.] And because AB is equal to CD, [Hyp.] and BC is common to the two triangles ABC, DCB; the two sides AB, BC are equal to the two sides DC, CB, each to each ; and the angle ABC was proved to be equal to the angle BCD; therefore the base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles, each to each, to which the equal sides are opposite ; [I. 4.] therefore the angle ACB is equal to the angle CBD.

And because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD. [I. 27.] And it was shewn to be equal to it.

Therefore the straight lines, &c. Q.E.D. Ex. If another parallelogram CEPD be described on CD, and on

the side remote from AB, then AB and AE shall be equal and parallel.

PROPOSITION 34. THEOREM.

The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them into two equal parts. 1.

Let ACBD be a parallelogram, of which BC is

a diameter; the opposite sides and angles of the figure shall be equal to one another, and the diameter BC shall bisect it. 3.

Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal

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to one another. [I. 29.) And because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal to one another. [I. 29.] Therefore the two triangles ABC, BCD have two angles ABC BCA in the one, equal to two angles DCB, CBD in the other, each to each, and one side BC is common to the two triangles, which is adjacent to their equal angles; therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other, namely the side AB equal to the side CD, and the side AC equal to the side BD, and the angle BAC equal to the angle CDB. [I. 26.]

And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD. [Ax. 2.] And the angle BAC has been shewn to be equal to the angle CDB. Therefore the opposite sides and angles of parallelograms are equal to one another.

Also their diameter bisects them. For AB being equal to CD, and BC common, the two sides AB, BC are equal to the two sides DC, CB, each to each; and the angle ABC has been proved equal to the angle BCD; therefore the triangle ABC is equal to the triangle BCD, [I. 4. and the diameter BC divides the parallelogram ACDB into two equal parts.

Therefore, the opposite sides, &c. Q.E.D.

Ex. 1. If one angle of a parallelogram be a right angle, all its

angles are right angles.
2. If the opposite sides of a quadrilateral figure be equal to

one another it is a parallelogram.
3. The diagonals of a parullelogram bisect each other.

PROPOSITION 35. THEOREM.

Parallelograms on the same base, and between the same parallels, are equal to one another. 1.

Let the parallelograms ABCD, EBCF be on the

same base BC, and between the same parallels AF, BC: the parallelogram ABCD shall be equal to the parallelogram EBCF. If the sides AD, DF A

E 3a. of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated in the same point D, it is plain what each of the parallelograms is double of the B triangle BDC; [I. 34.] and they are therefore equal to one another. [Ax. 6.]

E D

E

F

A

F

B

B

36.

But if the sides AD, EF, opposite to the base

BC of the parallelograms ABCD, EBCF be not terminated in the same point, then, because ABCD is a parallelogram AD is equal to BC; [I. 34.] for the same reason EF is equal to BC; therefore AD is equal to EF; [Ax. 1.] therefore the whole, or the remainder, AE is equal to the whole, or the remainder, DF. [Ax. 2, 3.] And AB is equal to DC'; [I. 34.] therefore the two EA, AB are equal to the two FD, DC each to each; and the exterior angle FDC is equal to the interior angle EAB; [I. 29.) therefore the base EB is equal to the base FC, and the triangle EAB is equal to the triangle FDC. [I. 4.] Take the triangle FDC from the trapezium ABCF, and from the same

trapezium take the triangle EAB, and the remainders are equal ; [Ax. 3.] that is, the parallelogram ABCD is equal to the parallelogram EBCF.

Therefore, parallelograms on the same base, &c. Q.E.D.

In the phrase the whole or remainder, which occurs in the demonstration, the whole refers to the left-hand figure, where AE is the sum of AD and DE, and the remainder to the right-hand figure, where AE is the difference of AD and DE.

At this point we enter upon the “third section” of the First Book, which treats of triangles and parallelograms upon equal bases, and between the same parallels and their relation to one another. Ex. 1. Make a rectangle which shall be equal in area to a given

parallelogram. 2. Make a rhombus which shall be equal in area to a given

parallelogram. 3. If a square and any other parallelogram of equal area

stand on the same base, the perimeter of the square shall be less than that of the parallelogram.

PROPOSITION 36. THEOREM. Parallelograms on equal bases, and between the same parallels, are equal to one another.

Let ABCD, EFGH be parallelograms on equal 1.

bases BC, FG, and between the same parallels AH, BG: the parallelogram ABCD shall be equal to the parallelogram EFGH. 2. Join BE, CH. А

E

F

B

Then, because BC is equal to FG, [Hyp.] and 3.

FG to EH, [I. 34.] BC is equal to E4; (Ax. 1.] and they are parallels, [Hyp.] 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. [I. 33.] Therefore BE, CH are both equal and parallel. Therefore EBCH is a parallelogram. [Def.] And it is equal to ABCD, because they are on the same base BC, and between the same parallels BC, AH. [I. 35.]

For the like reason the parallelogram EFGH is equal to the same EBBH. Therefore the parallelogram ABCD is equal to the parallelogram EFGH. [Ax. 1.]

Therefore, parallelograms, &c.

Q.E.D.

PROPOSITION 37. THEOREM.

2.

Triangles on the same base, and between the same parallels, are equal.

Let the triangles ABC, DBC be on the same 1.

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

Produce A DE

both ways to the points E, F; [ Post. 2.] through B draw BE parallel to CA, and through C draw CF parallel to BD. [I. 31.]

Then each of the figures EBCA, DBCF is a

parallelogram ; [Def.] and EBCA is equal to DBCF, because they are on the same base BC, and between the same parallels BC, EF. [I. 35.] And the triangle ABC is half of the parallelogram EBCA, because the diameter AB bisects it; [I. 34.] and the triangle DBC is half of the parallelogram DBCF, because the diameter DC bisects it. [I: 34.] But the

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