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THEOREM (Euclid 1. 34, First Part).

Repeat.-The enunciations of Euc. I. 26 and I. 29; and Axiom 2, and the definition of a parallelogram. General Enunciation. The opposite sides and angles of a paral

lelogram are equal to one another. Particular Enunciation.

Given.-The parallelogram A B C D.

B В

D

Required.—To prove that

(a) A B is equal to CD,
(b) A C is equal to BD,
(c) angle A B D is equal to angle A CD.

(d) angle B A C is equal to angle BDC. Construction.

Draw the diameter B C.

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Proof.

A B is parallel to CD (definition).
B C meets them.

by Euc. I. 29, (e) angle_A B C is equal to the alternate

angle BCD.

Again, A C is parallel to B D (definition),

B C meets them. .. by Euc. I. 29, (f) angle A CB is equal to the alternate

angle DBC. Then A B C, B C D are two triangles having the

two angles ABC, ACB equal to the two
angles BCD, DBC (e and f).
and the side B C common to the two triangles
and adjacent to the equal angles.

.. by Euc. I. 26, (g) the side A B is equal to the side CD, and (h) the side A C is equal to the side BD, and (i) the angle BAC is equal to the angle BDC. Again, angle A B C is equal to angle BCD (e), and angle D B C is equal to angle A CB (f).

by Axiom 2, (k) the whole angle ACD is equal to the whole angle A BD. Hence, by (8) and (h), The opposite sides are equal. And, by (i) and (k), The opposite angles are equal. Q. E. D.

TIIEOREM (Euclid I. 34, Second Part).

Repeat.-The enunciation of Euc. I. 4. General Enunciation.

The diameter of a parallelogram bisects it. Particular Enunciation. Given.--The parallelogram ABCD, and the

diameter B C (see next page). Required.-To prove that ABCD is bisected

by BC. Proof.

A B is equal to CD (g in last theorem),
to each of these add B C.

.. by Axiom 2a, (a) A B, B C are equal to DC, C B, each to each, and angle A B C is equal to angle BCD (e in last theorem).

by Euc. I. 4, The triangle A B C is equal to the triangle

BCD.

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But those two triangles make up the parallelogram

A B CD.
... ABCD is bisected by BC.

Q. E. D.
EXERCISE.
Given.-The parallelogram ABCD, and the

two diameters A D and B C.

A

B

D

Required. To prove that A D and B C bisect

each other in E. Use Eu. I. 34, first part, to prove A C and B D equal, and then I. 29 and 1. 26.)

THE EQUALITY OF GEOMETRICAL FIGURES.

In all cases where we have had to prove the equality of figures hitherto, those figures have been of the same form. That is to say, they have been similar in form as well as equal in area.

Thus in Euc. I. 4, we prove that the triangles are equal in area by superposition. This we could not do unless they were of exactly the same form.

But, of course, figures may be equal in area, though widely different in form. Thus a triangle and a square are of different form, and yet they may be of equal area; that is to say, they may contain the same number of square inches, feet, etc.

The succeeding propositions treat of figures which are equal in area, though not necessarily similar in form.

THEOREM (Euclid I. 35). Repeat.--The definition of a parallelogram. The enunciations of Euc. I. 4, I. 29, and I. 34. And Axioms 2a, 3a, and 6. General Enunciation. Parallelograms on the same base, and be

tween the same parallels, are equal to

one another. Particular Enunciation (CASE I).

A

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B

Given.-The parallelograms ABCD, DBCF,

on the same base B C, and between the same

parallels AF, BC, having the sides A D, DF

terminated at the same point D. Required.-To prove that ABCD, DBCF,

are equal. Proof.

ABCD is a parallelogram,
bisected by its diameter BD,
.. by Euc. I. 34,

triangle A B D is equal to triangle D BC. (a) And.. the parallelogram ABCD is double

of the triangle DBC. Again, DBCF is a parallelogram,

bisected by its diameter D C. .. by Euc. I. 34,

triangle FDC is equal to triangle D BC. (6) And.: the parallelogram DBCF is double

of the triangle DBC.
So that, comparing (a) and (6),

ABCD and DBCFare each double of DBC.
.: by Axiom 6,
ABCD is equal to DBCF.

Q. E. D.
Particular Enunciation (CASE II).
Given.—The parallelograms ABCD, EBCF,

on the same base B C; and between the same

D

B

parallels A F and BC. But having the side
E F apart from the side A D.
quired.—To prove that ABCD, EBCF,
re equal.

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