GEOMETRICAL AXIOMS

AXIOM 1. Magnitudes that can be made to coincide are

equal.

AXIOM 2. Through two points there can be made to pass one, and only one, straight line; and this may be indefinitely prolonged either way.

Hence,

a. Any straight line may be made to fall on any other straight line with any given point on the one on any given point on the other;

B. Two straight lines which meet in one point cannot meet again, unless they coincide.

AXIOM 3. Through the same point there cannot be more than one straight line parallel to a given straight line.

POSTULATES OF CONSTRUCTION

Let it be granted that

1. A straight line may be drawn from any one point to any other point.

2. A terminated straight line may be produced to any length in a straight line.

3. A circle may be drawn with any centre, with a radius equal

to any finite straight line.

BOOK II

EQUALITY OF AREAS

SECTION I

THEOREMS

DEF 1. The altitude of a parallelogram with reference to a given side as base is the perpendicular distance between the base

and the opposite side.

DEF 2. The altitude of a triangle with reference to a given side as base is the perpendicular distance between the base and the opposite vertex.

It follows from the General Axioms (d) and (e), as an extension of the Geometrical Axiom 1, that magnitudes which are either the sums or the differences of identically equal magnitudes are equal, although they may not be identically equal.

THEOR. 1. Parallelograms on the same base and between the same parallels are equal.

Let ABCD, EBCF be two parallelograms on the same base BC, and between the same parallels AF, BC:

then shall the parallelogram ABCD be equal to the parallelogram EBCF.

In the triangles DCF, ABE,

the angle FDC is equal to the angle EAB, since DC is parallel to AB,

I. 23, Cor. 2.

and the side CD is equal to the side BA,

the angle CFD is equal to the angle BEA, since CF is parallel to BE,

I. 23, Cor. 2.

I. 29.

I. 19.

since they are opposite sides of a parallelogram, therefore the triangles are identically equal.

From the quadrilateral ABCF take away the triangle DCF, and from the same quadrilateral take away the equal triangle АВЕ,

then the remainder ABCD is equal to the remainder EBCF.

Q.E.D. COR. 1. A parallelogram is equal to a rectangle, whose base and altitude are equal to those of the parallelogram.

For by the Theorem a parallelogram is equal to the rectangle on the same base, and of the same altitude, and by I. 30, Cor., this rectangle is equal to any other rectangle whose base and altitude are respectively equal to those of the other.

COR. 2. Parallelograms on equal bases and of equal altitudes are equal; and of parallelograms of equal altitudes, that

is the greater which has the greater base; and also of parallelograms on equal bases, that is the greater which has the greater altitude.

Parallelograms on equal bases and of equal altitudes are equal, since each parallelogram is equal to a rectangle whose base and altitude are equal to those of the parallelogram.

Again, rectangles of equal altitudes but on unequal bases may be superposed so that the rectangle which has the greater base is a whole of which the other rectangle is a part, and therefore less than the whole; hence, because a parallelogram is equal to a rectangle whose base and altitude are equal to those of the parallelogram, it follows that of parallelograms of equal altitudes that is the greater which has the greater base.

In like manner it may be shown that of parallelograms on equal bases, that is the greater which has the greater altitude.

Ex. 1. Prove Theor. I by I. 5, arranging the proof in three cases for the different positions of AD and EF.

Ex. 2. Prove Theor. I in the case in which D and E coincide, by the second part of I. 29.

Ex. 3. Shew how to divide any two parallelograms on the same base and between the same parallels into parts which are identically equal each to each.

THEOR. 2. A triangle is equal to half a rectangle whose base and altitude are equal to those of the triangle.

Let ABC be a triangle, and AD its altitude:

E

B

A

B

D

then shall the triangle ABC be equal to half a rectangle whose base is equal to BC and altitude to AD.

From B and C draw BE and CF parallel to AD, and let them meet EAF drawn through A parallel to BC in E and F.

Then EC is a rectangle whose base is BC and altitude AD.

Now the triangle ABD is half the rectangle ED, and the triangle ADC is half the rectangle AC,

I. 29.

I. 29.

therefore in fig. 1 the sum, or in fig. 2 the difference of the triangles ABD, ADC is half the sum or difference of the rectangles ED, AC,

therefore the triangle ABC is half the rectangle EC.

Q.E.D.

COR. 1. Triangles on the same or equal bases and of equal altitudes are equal.

For each triangle is half a rectangle whose base and altitude are equal to those of the triangles, and such rectangles are equal by I. 30, Cor.

COR. 2. Equal triangles on the same or equal bases have equal altitudes.

For triangles on the same or equal bases and of unequal altitudes are halves of rectangles on equal bases and of unequal