Def. A point taken in a finite straight line is said to divide the line internally, or simply, to divide it; and, if the line is produced, a point taken on the produced part, is said (by analogy) to divide the line externally; in either case the distances of the point from the extremities of the finite line are called the segments of the line.

Note-It follows, from this definition, that a straight line is equal to the sum or difference of its segments, according as it is divided internally or externally.

Propositions 9 and 10.

THEOREM—If a straight line is divided internally or externally at any point, the sum of the squares on the segments is double the sum of the squares on half the line and on the line between the point of division and the middle point of the line.

Let AB be a st. line whose mid pt. is M.

Let it be divided in X, internally fig. (1), or externally fig. (2). Then, by ii. 4, we have

sq. on AX = sq. on AM + sq. on MX + 2 rect. under AM, MX. And, by ii. 7, we have

sq. on BX + 2 rect. under BM, MX = sq. on BM + sq. on MX. , putting AM for BM, adding corresponding sides, and omitting 2 rect. AM, MX from each side, we get

sq. on AX + sq. on BX = 2 (sq. on AM + sq. on MX).

Note-It is clear that Props. 2 to 10 of Book ii are merely amplifications of Prop. I, and that they can be directly deduced from it. As a good exercise, the learner should make the deduction in each case.

Proposition 11.

PROBLEM-To divide a given straight line into two parts so that the rectangle contained by the whole and one part may be equal to the square on the other part.

Then the Ʌs at A being right, two sides of the sqs. must be in the same direction; and since AF < AB, a corner X of sq. on AF must be on AB.

Produce GX to meet DC in H.

Since AD is bisected in E and produced to F, .. rect. under DF, FA + sq. on AE = sq. on EF,

Def. When we speak of the projection of a terminated straight line on another straight line, we mean the distance between the feet of perpendiculars drawn from the extremities of the terminated line on the other.

NX

Note-If AP is a terminated line, and AX any other line through A; then, if PN is perpendicular to AX, AN is the projection of AP on AX.

Proposition 12.

THEOREM-In an obtuse-angled triangle the square on the side opposite the obtuse-angle is greater than the sum of the squares on the sides forming the obtuse angle, by twice the rectangle contained by either of these latter sides, and the projection of the other upon it.

A

C

N

Let ABC be a ▲ in which ACB is obtuse.

Draw AN to BC produced; so that CN is the projection of AC on BC.

Then, by ii. 4, we have

sq. on BN = sq. on BC + sq. on CN + 2 rect. under BC, CN. .., adding sq. on AN to each side; and recollecting that

sq. on BN + sq. on AN = sq. on AB,

sq. on AB

sq. on AC,

N is right,

=sq. on BC + sq. on AC + 2 rect. under BC, CN.

Proposition 13.

THEOREM-In any triangle the square on a side opposite an acute angle is less than the sum of the squares on the sides forming the acute angle, by twice the rectangle contained by one of the latter lines and the projection of the other upon it.

(1)

(2)

A A

N

Let ABC be a ▲ in which B is acute.

Draw AN to BC, which may need to be produced, as in fig. (1), or not, as in fig. (2): then BN is the projection of AB on BC.

.., in both cases, by ii. 7, we have

sq. on BC + sq. on BN = sq. on CN + 2 rect. under BC, BN.

.., adding sq. on AN to each side; and recollecting that

sq. on BN + sq. on AN = sq. on AB, and sq. on CN + sq. on AN = sq. on AC, }

we get

}

N is right,

sq. on BC + sq. on AB= sq. on AC + 2 rect. under BC, BN:

i. e.

sq. on AC < sq. on BC + sq. on AB by 2 rect. under BC, BN.

H

Proposition 14.

PROBLEM-To describe a square whose area shall be equal to the area of a given rectilineal figure.

A

Let A be the given rectil. fig.

Make a rect. BCDE, which = A.

Then, if BE BC, BD is a sq., and problem is solved.

=

Assume BE and BC unequal.

Produce BE to F, so that EF = ED.

Bisect BF in O; and with O as centre, and BO as radius, describe a O.

Produce DE to meet this in X; and join OX.

Since BF is divided equally in O, and unequally in E, .. rect. under BE, EF + sq. on OE = sq. on BO,

= sq. on OX,

= sq. on EX + sq. on OE.

Take sq. on OE from each side.

Then rect. under BE, EF = sq. on EX:

i. e. the sq. described on EX =

fig. A.