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THEOREM (Euclid I. 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 parallelogram are equal to one another.
Given. The parallelogram A B C D.
Required. To prove that (a) A B is equal to CD, (b) A C is equal to BD, (c) angle ABD is equal to angle A CD. (d) angle BAC is equal to angle B D C. Construction.
Draw the diameter B C.
AB is parallel to CD (definition).
BC meets them.
.. by Euc. I. 29,
(e) angle ABC is equal to the alternate angle B C D.
Again, AC is parallel to BD (definition),
by Euc. I. 29,
(f) angle A CB is equal to the alternate angle D B C.
Then ABC, BCD are two triangles having the two angles ABC, ACB equal to the two angles B CD, DBC (e and f).
and the side B C common to the two triangles
(g) the side A B is equal to the side C D, and (h) the side A C is equal to the side B D, and (i) the angle BAC is equal to the angle BDC. Again, angle A B C is equal to angle B CD (e), and angle D B C is equal to angle A CB (ƒ). .. by Axiom 2,
(k) the whole angle A CD is equal to the whole angle A B D.
Hence, by (g) and (h),
The opposite sides are equal.
The opposite angles are equal.
Q. E. D.
THEOREM (Euclid I. 34, Second Part).
Repeat. The enunciation of Euc. I. 4.
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.
AB is equal to CD (g in last theorem),
.. by Axiom 2a,
(a) AB, BC are equal to DC, C B, each to each, and angle ABC is equal to angle B CD (e in last theorem).
by Euc. I. 4,
The triangle A B C is equal to the triangle BCD.
But those two triangles make up the parallelogram
A B CD.
... A B C D is bisected by BC.
Given. The parallelogram ABCD, and the two diameters A D and BC.
Q. E. D.
Required. To prove that AD and BC bisect each other in E. Use Euc. I. 34, first part, to prove AC and BD equal, and then I. 29 and I. 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.
Parallelograms on the same base, and between the same parallels, are equal to one another.
Particular Enunciation (Case I).
Given. The parallelograms ABCD, DBCF, on the same base BC, and between the same
parallels AF, BC, having the sides AD, DF terminated at the same point D.
Required. To prove that ABCD, DB CF, are equal.
ABCD is a parallelogram,
triangle ABD is equal to triangle DB C. (a) And.. the parallelogram ABCD is double of the triangle DBC.
Again, DB CF is a parallelogram,
bisected by its diameter D C.
.. by Euc. I. 34,
triangle FDC is equal to triangle DB C. (b) And.. the parallelogram DBCF is double of the triangle DBC.
So that, comparing (a) and (b),
A B CD and DBCF are each double of DBC. .. by Axiom 6,
ABCD is equal to DBCF.
Q. E. D.
Particular Enunciation (CASE II). Given. The parallelograms A B C D, E B CF, on the same base BC; and between the same
parallels AF and B C. EF apart from the side Required. To prove that ABCD, EBCF, are equal.
But having the side AD.