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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 (6) 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 A D) is not parallel to B C
.. AD is parallel to B C.
I. EXERCISE. (Euc. I. 40.)
Q. E. D.
Given. The triangles A B C, DEF equal to
one another; on equal bases B C, E F; 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. I. 38.)
II. EXERCISE. (Euclid I. 41)
Given. The parallelogram ABCD and the triangle EB C, on the same base B C, and between the same parallels A E and B C.
Required. To prove that A B CD is double of
(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.
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.
(a) Bisect BC at E (by Euc. I. 10). Join AE (see next page).
At the point E, in the straight line E C,
(b) Make the angle CEF equal to angle D.
(c) Through A draw A FG parallel to BC, and
(d) Through C draw C G parallel to E F (by Euc. I. 31).
By (c) and (d), and definition,
.. by Euc. I. 38,
The triangle ABE is equal to the triangle AEC.
(e). The triangle ABC is double of the triangle A E C.
Again, the parallelogram FE CG 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 (ƒ), and Axiom 6,
its angle F E C is equal to angle D. Wherefore, a parallelogram FECG has been described equal to the given triangle A B C, and having one of its angles FEC equal to the given angle D.
Q. E. F.
in the same straight line
(Produce AB, making BD equal to AB, and through C draw CE parallel to AD. 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.
The complements of the parallelograms which are about the diameter of a parallelogram, are equal to one another."
EXPLANATION. The parallelograms AEKH,
KGCF are said to be described about the diameter
And the remaining parts of the whole parallelogram —that is, the parallelogram E B G K, and the parallelogram HKFD-are called the complements of the parallelograms about the diameter.
Given. The parallelogram A B C D, having the parallelograms AEKH, KGCF about the diameter A C.
Required. To prove that EB GK is equal to HKFD.
ABCD is a parallelogram, and A C its diameter. .. by Euc. I. 34,
(a) the triangle ABC is equal to the triangle A D C.
A E KH is a parallelogram, and A K its diameter. ... by Euc. I. 34,
(b) the triangle A E K is equal to the triangle AH K.
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 AEK, KG C, from the triangle ABC, the remainder is the parallelogram E B GK.
Take away the triangles AHK, KFC from the triangle ADC, the remainder is the parallelogram HKFD.
Hence, by (a), (b), and (c), and Axiom 6,
EBGK is equal to HKF D.
Q. E. D.