Imágenes de páginas
PDF
EPUB

Ax. 6.

[blocks in formation]

FB, GC.

and square GB = 2 ▲ FBC and between

But the doubles of equals are equal,

BL = square GB.

In the same way, by joining AE, BK, it can

CL

be proved that

square HC.

[ocr errors][merged small]

whole square
BDEC

=square GB+square HC,

[blocks in formation]

Wherefore, in any right-angled triangle, &c.

[blocks in formation]

If one side of a triangle be produced, and the exterior angle, also one of the interior and opposite angles, be each of them bisected; the remaining interior and opposite angle will be double the angle made by the bisecting lines.

Let CBA be a A. Prod. BC to D, and let the ext. ACD be bisected by the str. line CE, and the int. and opp. ABC by the str. line BE. Then will BAC = 2▲ BEC.

[blocks in formation]

1st. The lines CE, BE, will meet in some

point, as E. For

LACD> <ABC,

ACD> ABC,

Prop. 15.

i. e.

.. but

<ECD> <EBC:

add to each the ECB.

ECD+ ECB > < EBC + / ECB;
ECD+ ▲ ECB= 2 rt. s,

ZEBC+ ECB < 2 rt.

S.

Prop. 12.

.. CE, BE, will meet on the same side of Ax. 12. the str. line BC on which the ABC is.

CBE + BEC,

2ndly. In BCE, the int. and opp.s, Prop.31.

ext. DCE

..2/DCE or / DCA = 2 / CBE+2 / BEC·

ZABC = = 2/ CBE,

but

DCA =

In Δ ABC,

ZABC+2 BEC.

ABC+BAC; Prop.31.

ext.DCA =

.. ZABC+▲ BAC = ZABC+2 BEC. Ax. 1. from these equals take ▲ ABC

BAC

[ocr errors]

2/ BEC.

Wherefore, if one side of a triangle, &c.

Note. This proposition is introduced on account of the explanation that it affords of the principle upon which the sextant depends.

Ax. 3.

DEFINITIONS.

ང་

EVERY right-angled parallelogram, or rectangle, is said to be contained by any two of the straight lines which contain one of the right angles.

II.

In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a Gnomon. 'Thus the parallelogram HG, together with the complements AF, FC is the Gnomon; which is more briefly expressed by the letters AGK or EHC, which are at the opposite angles of the parallelograms which make the gnomon.'

[blocks in formation]

N.B.-The rectangle contained by any two straight lines AB, AD, is generally expressed by the words "rectangle of AB and AD;" or by "AB. AD.”

PROP. XLI.

THEOR.

1. 2 Eu.

If there be two straight lines, one of which is divided into any number of parts; the rect

angle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line.

Let A and BC be two str. lines, and let BC be divided into any parts in D and E; then A.BC=A. BD+A.DE+A. EC

[blocks in formation]

BF BC,
BG= A;

draw

GH || BC; and DK, EL, CH || BG.

[merged small][merged small][ocr errors][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small]

.. A. BC=A. BD+A.DE+A.EC. Wherefore, if there be two str. lines, &c.

PROP. XLII. THEOR. 3. 2 Eu.

If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rect

Prop. 10.
Prop. 3.

Prop. 30.

Prop. 33.

angle contained by the two parts, together with the square of the aforesaid part.

Let the str. line AB be divided into two parts in the pt. C; then AB. BC= AC. BC +BC2.

[blocks in formation]

.. AB. BC= AC. BC + BC3.

Therefore, if a str. line be, &c.

Note. If the parts of the divided line be a and b; the proposition algebraically expressed, is a+b.a=ab+a2.

PROP. XLIII. THEOR. 4. 2 Eu.

If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.

« AnteriorContinuar »