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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 im

possible. Therefore, the supposition (6) on which (d) is

founded is impossible,

and A E is not parallel to BC. In the same way it could be shown that every line

drawn through A (except A D) is not parallel

to BC. .. A D is parallel to B C.

Q. E. D. I. EXERCISE. (Euc. I. 40.) Given.- The triangles ABC, DEF equal to

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one another; on equal bases B C, EF; in the same straight line B F, 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 B F, draw another line meeting D E, and suppose that line parallel, and use Euc. I. 38.)

II. EXERCISE. (Euclid I. 41) Given.--The parallelogram ABCD and the

triangle E B C, on the same base B C, and between the same parallels A E and B C.

Required.—To prove that A B C D is double of

Е В С.

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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.

iven.-The triangle ABC, and the angle D. Required. - To describe a parallelogram equal to ABC, and having one of its an. gles equal to D.

B Construction. (a) Bisect B C at E (by Euc. I. 10).

Join A E (see next page).

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K

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

B

(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 BC).

by Euc. I. 38, The triangle A B E is equal to the triangle

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

triangle AEC. Again, the parallelogram FECG and the triangle A E C are on the same base, and between the same parallels. :. by Euc. I.

41, (f) FECG is double of A E C. Hence, by (e) and (f), and Axiom 6,

FECG is equal to ABC, and

its angle FEC is equal to angle D. Wherefore, a parallelogram FECG has been de

scribed equal to the given triangle A B C, and having one of its angles FEC equal to the given angle D.

Q. E. F.

EXERCISE.
Given.-The triangle A B C.
Required.-To construct

another triangle, equal to
A B C, having its base
in the same straight line
with A B, and its vertex
in a line drawn through

C parallel to A B. (Produce AB, making B D equal to AB, and through C draw CE parallel to A D. Describe a triangle on B D 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 paral

lelogram, are equal to one another. EXPLANATION. - The parallelograms A EKH,

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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 E B GK, and the parallelogram H KFD-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 C. Required.—To prove that EBGK is equal to

HKFD. Proof. A B C D is a parallelogram, and A C its diameter.

.. by Euc. I. 34, (a) the triangle ABC is equal to the tri

angle ADC. A E K H is a parallelogram, and A K its diameter.

.. by Euc. I. 34, (b) the triangle A EK is equal to the tri

angle A HK. KGCF is a parallelogram, and KC its diameter.

... by Euc. I. 34, (c) the triangle KGC is equal to the tri

angle KFC. Take away the triangles A EK, KG C, from the

triangle A B C, the remainder is the parallel

ogram EBG K. Take away the triangles AHK, KFC from the

triangle ADC, the remainder is the parallel

ogram H KFD. Hence, by (a), (6), and (c), and Axiom 6, E BG K is equal to HKFD.

Q. E, D.

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