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.

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

THEOREM I.

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

Part. En. Let ABCD, EBCF be parallelograms, upon the same base BC, and between the same parallels AF, BC;

it is required to prove that the parallelogram ABCD is equal to the parallelogram EBCF.

Proof. Because ABCD is a parallelogram,

therefore

AB = DC;

(Hyp.)

(I. 28.)

because AB, BE are respectively parallel to CD, CF, (Hyp.)

therefore the angles at A and E are respectively equal to the corresponding angles at D and F;

therefore the triangles ABE, DCF are equal.

(1. 23, Cor.)

(I. 17.)

But if the triangle CDF is taken away from the trapezium ABCF the parallelogram ABCD remains; and if the triangle ABE is taken away from the same trapezium the parallelogram EBCF remains; therefore the parallelogram ABCD is equal to the parallelogram EBCF*.

COR. I. The area of a parallelogram is equal to the area of a rectangle, whose base and altitude are equal to those of the parallelogram.

COR. 2. Parallelograms on equal bases and of equal altitude 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.

THEOREM 2.

The area of a triangle is half the area of a rectangle whose base and altitude are equal to those of the triangle.

Let ABC be a triangle on the base AC, and DACE the rectangle on the same base, and having the same altitude as the triangle;

then will the area of the triangle ABC be half that of the rectangle DACE.

Constr. Through C draw CF parallel to AB, to meet DE produced in F

Then BACF is a parallelogram, and therefore BACF is equal to the rectangle DACE.

(II. I.)

But the triangle ABC is half the parallelogram BACF;

(I. 28.)

therefore the triangle ABC is half the rectangle DACE.

COR. I.

Triangles on the same or equal bases and of equal altitude are equal".

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

COR. 3. If two equal triangles stand on the same base and on the same side of it, or on equal bases in the same straight line and on the same side of that straight line, the line joining their vertices is parallel to the base or to that straight linet.

THEOREM 3.

The area of a trapezium is equal to the area of a rectangle whose base is half the sum of the two parallel sides, and whose altitude is the perpendicular distance between them.

Part. En. Let ABCD be a trapezium having AD parallel to BC;

then it is required to prove

* Euclid, I. 37, 38.

that its area is equal to

+ Euclid, I. 39, 40.

that of the rectangle whose base is half the sum of AD and BC, and altitude the perpendicular distance between AD and BC.

E C

Proof. Bisect DC in O, and through O draw a line parallel to AB to meet BC in E, and AD produced

(1. 7.)

therefore the triangles are equal in all respects;

and therefore the trapezium ABCD is equal to the parallelogram ABEF.

But the parallelogram ABEF is equal to the rectangle on the same base BE, and between the same parallels;

(II. 1. Cor.)

and since

and

EC=DF,
AF= BE,

therefore the base BE is half the sum of AD and BC;

therefore the trapezium ABCD is equal to the rectangle whose base is half the sum of the parallel sides, and height the perpendicular distance between them.

Def. 3. The straight lines drawn through any point in a diagonal of a parallelogram parallel to the sides divide it into four parallelograms, of which the two whose diagonals are upon the given diagonal are called parallelograms about that diagonal, and the other two are called the complements of the parallelograms about the diagonal.

THEOREM 4.

The complements of parallelograms about the diagonal of any parallelogram are equal to one another.

it is required to prove that the complement PB = the complement PD.

Proof. For the triangle ABC=the triangle ADC (1. 28); and the triangles ASP, PQC=the triangles ARP, PTC; therefore the remainders are equal, that is, PB = PD*.

Def. 4. All rectangles being identically equal which have two adjoining sides equal to two given straight lines, any such rectangle is spoken of as the rectangle contained by

those lines.

In like manner, any square whose side is equal to a given straight line is spoken of as the square on that line.

Def. 5. A point in a straight line is said to divide it internally, or, simply, to divide it; and, by analogy, a point

* Euclid, I. 43.