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[blocks in formation]

.. square on D + rectangle D, B = rectangle D, A ; (11. 1) But rectangle D, A + rectangle A, B = square on A, (11.1) and rectangle D, B+ square on B = rectangle A, B, (11. 1) .. square on D + rectangle D, B+ rectangle A, B

= rectangle D, A + rectangle A, B,

..=square on A;

.. sq. on D+rect. D, B+sq. on B+rect. A, B=sqq. on A, B; ... square on D + twice rectangle A, B

=

squares on A,

B.

COR. It follows from the proposition that

squares on S, D+ twice rectangle A, B

=

twice squares on A, B + twice rectangle A, B; .. squares on S, D= twice squares on A, B.

Also that

square on S = squares on A, B + twice rectangle A,
and
square on D + four times rectangle A, B.

N.B.

B,

By the method here adopted all propositions which are the Geometrical interpretations of Algebraical identities of the second degree may be established.

DEFINITIONS.

An obtuse angle is one greater than a right angle.
An acute angle is one less than a right angle.

An obtuse-angled triangle is one having an obtuse angle.

An acute-angled triangle is one having three acute angles.

PROPOSITION III.

In obtuse-angled triangles, the square on the side opposite the obtuse angle is greater than the squares on the sides containing it by twice the rectangle contained by either of those sides and the part of it intercepted between the perpendicular let fall upon it from the opposite angle and the obtuse angle.

[blocks in formation]

Let ABC be an obtuse-angled ▲, having the obtuse LACB,

and from B let fall BK to AC produced. Then shall the square on AB be > the squares on AC, CB by twice the rectangle AC, CK.

For AK is the sum of AC and CK;

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.. square on AK is = squares on AC, CK

+ twice rectangle AC, CK; (11.2)

.. squares on AK, KB are = squares on AC, CK, KB

+ twice rectangle AC, CK;

but squares on AK, KB are

and squares on CK, KB are

square on AB,

square on CB ;

.. square on AB is = squares on AC, CB

+ twice rectangle AC, CK.

PROPOSITION IV.

In any triangle the square on the side opposite to any one of the acute angles is less than the squares on the sides containing it by twice the rectangle contained by either of those sides and the part of it intercepted between the perpendicular let fall upon it from the opposite angle and the acute angle.

AAA

Let ABC be a ▲ having the acute 4 ACB, and from B let fall BK 1 CA or CA produced. Then shall the square on AB be < the squares on AC, CB by twice the rectangle AC, CK.

For should the

fall within or without the ▲ ABC, in either case AK is the difference between AC and CK.

=

.. square on AK + twice rectangle AC, CK is = squares on AC, CK.

(II. 2)

.. squares on AK, KB + twice rectangle AC, CK= squares on AC, CK, KB.

.. square on AB + twice rectangle AC, CK = squares on AC, CB.

Should the BAC be a right 4, and ... BK coincide with BA,

then square on AB + twice square on AC = squares on АС, СВ.

(1.35)

PROBLEM A.

Describe a square equal to a given rectilineal figure.

[blocks in formation]

Let A be the given rectilineal figure. First describe the rectangle FH = A ; then if the sides are equal FH is a square, and what was required is done.

M

But, if not, produce FG and cut off GM = GH.
Bisect FM in P,

and with centre P and radius PM describe ✪ FQM. Produce HG to meet the O in Q.

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.. rectangle FG, GM is the difference of the squares on PM and PG,

and is ...

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(II. 2)

the difference of the squares on PQ and PG, = the square on GQ,

and is ...

but FH is

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(I. 35)

... square on GQ = A.

PROBLEM B.

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

Let PQ be the given straight line.
On PQ describe the square PR.

Bisect PS in K and join QK. Produce SP and cut off KG = KQ. On PG describe a square GPCF, one side of which PC will fall on PQ. Then shall PQ be divided in C, so that rectangle PQ, QC= square on PC.

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K

H

[blocks in formation]

... rectangle SG, GP is = the difference of the squares

on KG and KP,

and is ... =

(II. 2)

the difference of the squares on KQ and KP, and is ... = the square on PQ;

(1.35)

but figure GH = rectangle SG, GP; for GF= GP,
and figure PR is the square on PQ;

.. figure GH = figure PR.

Take away the common part, the figure PH,

... the remainder GC

=

the remainder CR,

but GC is the square on PC and CR is = rectangle PQ, QC, for QR is PQ;

=

... rectangle PQ, QC = square on PC.

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