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.

A

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. I. 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 rectangle contained by the two parts.

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‡.

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 S.

Part. En. Let AB be the sum 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.

Proof. Describe a square ADEB on AB.

A

P

B

Through P draw PLF parallel to AD, meeting DE

in F: cut off PL= PB leaving LF AP. Through Z draw

HLM parallel to AB, to meet DA and EB in M, H.

營 Euclid, II. I. + Euclid, II. 3.

Euclid, II. 2. § Euclid, II. 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 rectangle AP, PB.

Proof. Describe a square ADEB A on AB.

M

D

H L

B

E

F

Through P draw LPF parallel to AD, meeting DE produced in F: cut off PL=PB, making _LF=AP. Through Z 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, II. 7.

And AE is less than MF+ HP by MP + HF; that is, the square on AB is equal to the squares of AP, PB diminished by twice the rectangle contained by AP, PB.

THEOREM 8.

The difference of the squares on two lines is equal to the rectangle contained by the sum and difference of the lines*.

Part. En. Let AB and BC be the two straight lines, of which AB is the greater; and let them be placed in one straight line; cut off BD equal to BC; so that AC is their sum, and AD is their difference;

Then will the difference of the squares of AB and BC be equal to the rectangle contained by AC and AD.

difference of the squares of AB and BC is the figure made up of EL and AO.

But EL is equal to BF by construction;

therefore the figure made of EL and AO is equal to AF; which is the rectangle contained by AE or AD and

AC;

Euclid, II. 5, Cor.

therefore the difference of the squares of AB and BC is equal to the rectangle contained by AC and AD.

COR. If a straight line is bisected and divided in any point, the rectangle contained by the segments is equal to the difference of the squares on half the line and the line between the points of section.

Proof. For let AB be bisected in C, and divided internally or externally in P.

Then AP is the sum of AC and CP, and PB is their difference, since BC= AC.

Therefore the rectangle contained by AP, PB is the rectangle contained by the sum and difference of AC and CP, and therefore is equal to the difference of the squares of AC and CP.

Remark. The student will begin here to suspect, what he will afterwards find to be true, that there is an intimate relation between geometry and algebra. Algebraical or analytical geometry as it is called, investigates this relation and applies it to the establishment of theorems in geometry, and will occupy him at a later stage of his mathematical studies. We shall at present use the expression AB3, which is read 'AB squared,' only as an abbreviation for "the square on AB," and ABX AC or AB. AC, as an abbreviation for "the rectangle contained by AB and AC."

THEOREM 9.

In any right-angled triangle the square on the hypotenuse is equal to the sum of the squares on the sides which contain the right angle.