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But triangle A B C is equal to triangle D B C (a). ... by Axiom 1,

(d) triangle EBC is equal to triangle DBC. The less equal to the greater, which is impossible.

Therefore, the supposition (b) on which (d) is founded is impossible,

and AE is not parallel to BC.

In the same way it could be shown that every line drawn through A (except AD) is not parallel to BC.

... AD is parallel to BВ С.

I. EXERCISE. (Euc. I. 40.)

Q. E. D.

Given. The triangles ABC, DEF equal to

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one another; on equal bases BC, EF; in the same straight line BF, and on the same side of it.

Required. To prove that ABC, DEF are between the same parallels.

(Proceed, as in last theorem, by supposing the line joining A D not parallel to BF, draw another line meeting DE, and suppose that line parallel, and use Euc. Ι. 38.)

II. EXERCISE. (Euclid I. 41)

Given. The parallelogram ABCD and the triangle EBC, on the same base BC, and between the same parallels A E and B C.

Required. To prove that A B CD is double of

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(Join A C. Then you have two triangles on the same base, and between the same parallels. Use Euc. I. 37 and I. 34.)

PROBLEM (Euclid I. 42).

Repeat. The enunciations of Euc. I. 38, I. 41, and Axiom 6.

General Enunciation.

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

Particular Enunciation.

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(a) Bisect BC at E (by Euc. I. 10). Join A E (see next page).

K

C

At the point E, in the straight line E C, (6) Make the angle CEF equal to angle D.

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(c) Through A draw AFG parallel to BC, and

(d) Through C draw CG parallel to E F (by Euc. I. 31)..

Proof.

By (c) and (d), and definition,

FECG is a parallelogram.
BE is equal to EC (a), and
AG is parallel to B C (c).

... by Euc. I. 38,

The triangle ABE is equal to the triangle

AEC.

(e) ... The triangle ABC is double of the triangle AEC.

Again, the parallelogram FECG and the triangle AEC are on the same base, and between the same parallels.

... by Euc. I. 41,

(f) FECG is double of AE C. Hence, by (e) and (f), and Axiom 6, FECG is equal to ABC, and

its angle FE C is equal to angle D.

Wherefore, a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles FEC equal to the given angle D.

Q. E. F.

EXERCISE.

Given. The triangle A B С.
Required. To construct
another triangle, equal to
ABC, having its base
in the same straight line
with A B, and its vertex
in a line drawn through

A

B

C parallel to A В.

(Produce AB, making BD equal to AB, and through C draw CE parallel to A D. Describe a triangle on BD having its vertex in CE; and use Euc. I. 38 for proof.)

THEOREM (Euclid I. 43).

Repeat. The enunciation of Euc. I. 34, and Axioms 2 and 3. General Enunciation.

The complements of the parallelograms which are about the diameter of a parallelogram, are equal to one another.

EXPLANATION. The parallelograms AEK Η,

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KGCF are said to be described about the diameter A C.

And the remaining parts of the whole parallelogram -that is, the parallelogram EBGK, and the parallelogram HKFD-are called the complements of the parallelograms about the diameter.

Particular Enunciation.

Given. The parallelogram ABCD, having the parallelograms AEKH, KGCF, about the diameter A С.

Required. To prove that EBGK is equal to HKFD.

Proof.

A B C D is a parallelogram, and A Cits diameter. ... by Euc. I. 34,

(a) the triangle ABC is equal to the triangle ADC.

AEKH is a parallelogram, and A K its diameter. ... by Euc. I. 34,

(b) the triangle AEK is equal to the triangle A Η Κ.

KGCF is a parallelogram, and KC its diameter. ... by Euc. I. 34,

(c) the triangle KGC is equal to the triangle KFC.

Take away the triangles A EK, KGC, from the triangle A B C, the remainder is the parallelogram EBGK.

Take away the triangles A HK, KFC from the triangle ADC, the remainder is the parallelogram HKF D.

Hence, by (a), (b), and (c), and Axiom 6,
EBGK is equal to HKFD.

Q. E. D.

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