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Ax. 1.

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(and its opp. sides are equal;
AB = AD,


= AD=DE=
and ADEB is equilat.

Also AD meets the ||s AB, DE,


Prop. 28.

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<BAD+▲ ADE = 2 rt. s

Constr. Ax. 3. Prop. 33. Ax. 1.

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thes ABE, BED, are rt. Ls,

..the figure ADEB is rectangular and equilat. Def. 30... the figure is a square, and descr. on AB.

all its angles rt. s.

TM that has one rt. ▲, has


THEOR. 47.1 Eu.

In any right-angled triangle, the square which
is described upon the hypothenuse, or side
subtending the right angle, is equal to the
sum of the squares described upon the sides
which contain the right angle.

Let ABC be a rt. angled, having the rt.
BAC. Then BC BA2 + AC2.


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On BC, BA, AC, descr. the squares BE, Prop. 38.

GB, HC.*

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= 2 rt. s.

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CA is in the same str. line with AG.

For the same reason

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AB is in the same str. line with AH,

and DBC =

FBA, ea. being a rt. • Def. 30.

Add to each / ABC;

.. whole DBA = whole / FBC, AB, BD = FB, BC, ea. to ea. ▲ DBA = ▲ FBC,


Ax. 2.

Def. 30.

Prop. 4.

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Again, BL = 2 ABD and between Prop.35.

Cor. 2.

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Squares and parallelograms are frequently expressed, for the sake of brevity, by the letters at their opposite angles; and the square on any line, as BC, is represented by BC2.

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In the same way, by joining AE, BK, it can be proved that

whole square

Ax. 2.

i. e.


square HC.

BDEC = square GB+square HC,
BC2= BA2+AC2.

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

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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. BAC2 BEC.

Then will

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

point, as E. For

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EBC + ≤ ECB;

Prop. 12.


but ECD+ ▲ ECB= 2 rt. Zs, EBC+ ECB < 2 rt. .. CE, BE, will meet on the the str. line BC on which the

2ndly. In



same side of Ax. 12. ABC is.

the int. and opp. / s, Prop.31.

.. 24 DCE or DCA = : 2 / CBE+2 / BEC ·

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▲ ABC+ ▲ BAC; Prop.31.

BEC. Ax. 1.

.. ZABC+2 BAC = ZABC+2

from these equals take ABC
BAC =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.



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.


In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a 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.'

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


THEOR. 1. 2 Eu.

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

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