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

Proof. Bisect DC in 0, and through o draw a line parallel to AB to meet BC in E, and AD produced in F.

Then in the triangles DOF, EOC, because DO = OC,

(Constr.) and the angle DOF= the angle EOC,

(1. 4.) } and the angle ODF= the angle OCE;

(1. 22.)) therefore the triangles are equal in all respects; (1. 7.) 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;

(11. I. Cor.) and since

EC=DF, and

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.

Let ABCD be a parallelogram, P any point on the diagonal AC, and let RPQ, SPT be drawn parallel to the sides; 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, 1. 43.

in the line produced is said to divide it externally; and, in either case, the distances of the point from the extremities of the line are called its segments.

Obs. A straight line is equal to the sum or difference of its segments according as it is divided internally or externally.

THEOREM 5.

The rectangle contained by two given lines is equal to the sum of the rectangles contained by one of them and the several parts into which the other is divided.

Part. En. Let A and BC be the two given lines, of which BC is divided into any number of parts, BD,DE,EC; it is required to prove that the rectangle contained by A and BC is equal to the sum of the rectangles contained by A and BD, A and DE, A and EC.

Proof. From B draw a line BF at right angles to BC, and equal to A; through F draw a line parallel to BC; and through D, E, C draw DG, EH, CK parallel to BF.

Then the figure BK is equal to the figures BG, DH, EK: but BK is the rectangle contained by A and BC; and BG, DH, EK are respectively the rectangles contained by A and BD, A and DE, A and EC;

therefore the rectangle contained by A and BC is equal to the rectangles contained by A and BD, A and DE, A and EC*.

Cor. 1. If a straight line is divided into two parts, the rectangle contained by the whole line and one of the parts is equal to the sum of the square on that part and the rectang!e contained by the two parts t.

CoR. 2. If a straight line is divided into two parts the square on the whole line is equal to the sum of the rectangles contained by the whole line and each of the parts I.

THEOREM 6.

The square on the sum of two lines is greater than the sum of the squares on those lines by twice the rectangle contained by them . Part. En. Let AB be the sum

p. of AP, PB;

it is required to prove that the square on AB is equal to the squares on AP, PB together with twice the rectangle contained by AP, PB.

F E Proof. Describe a square ADEB on AB.

Through P draw PLF parallel to AD, meeting DE in F: cut off PL= PB leaving LF= AP. Through I draw HLM parallel to AB, to meet DA and EB in M, H. * Euclid, 11. 1. + Euclid, 11. 3. Euclid, 11. 2. § Euclid, 11. 4.

Then the figures AL, PH, LE, MF are rectangles by construction; and PH, MF are the squares on PB, AP respectively; and AL, LE are each of them the rectangle contained by AP, PB.

Hence, since ADEB is made up of these four figures, it follows that the square on AB is greater than the squares on AP, PB by twice the rectangle contained by AP, PB.

THEOREM 7.

The square on the difference of two lines is less than the sum of the squares on those lines by twice the rectangle contained by them*.

Part. En. Let AB be the difference of AP, BP; it is required to prove that the square on AB is less than the squares on AP, PB by twice the

M

TIL rectangle AP, PB.

Proof. Describe a square A DEB on AB.

Through P draw LPF parallel to AD, meeting DE produced in F: cut off PL= PB, making LF=AP. D Through I draw MHL parallel to AB, to meet DA and EB produced in M, H.

Then MF is the square on AP; and HP the square on BP; and MP or HF, the rectangle contained by AP and BP.

* Euclid, 11. 7.

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