Imágenes de páginas
[merged small][ocr errors][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small]


If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and opposite angle on the same side; and likewise the two interior angles upon the same side together equal to two right angles.

Let the str. line EF fall on the || str. lines AB, CD, then

alt. / AGH = alt. / GHD,

ext. EGB = int. / GHD,
And BGH+▲ GHD = 2 rt. s.

[blocks in formation]

Prop. 12. but

AGH+ / BGH= 2 rt.

≤ s,

[ocr errors]


Ax. 12.



BGH+≤ GHD<2 rt.

AB, CD, will meet if prod. far enough; but they never meet, since they are parallel;


LAGH is not

Prop. 14.

i. e. but

< AGH=/ EGB,

Ax. 1.

[ocr errors]


Ax. 2.


add to each the BGH,


Prop. 12. but


EGB+ / BGH = 2 rt. s,

BGH+ ≤ GHD= 2 rt. s.

Wherefore, if a straight line, &c.

PROP. XXIX. THEOR. 30. 1 Eu.

Straight lines which are parallel to the same straight line, are parallel to each other.

Let AB, CD, be each || EF; then shall

[blocks in formation]

Let str. line GK cut AB, EF, CD, in the points G, H, K.

Then GK cuts the lines AB, EF,

Prop. 28.

Again, ... GK cuts the || lines EF, CD,
GHF: = / GKD,

Prop. 28.


Ax. 1.

and it was proved / AGH = / GHF

/ AGH = / GKD,

which are alt. s.

... AB | CD.

Wherefore str. lines, &c.

PROP. XXX. PROB. 31. 1 Eu.

To draw a straight line through a given point, parallel to a given straight line.

Let A be the given point, BC the given str. line. To draw a str. line through A || BC. In BC take any pt. D,

join AD,

make DAE =

prod. EA to F,


then shall EF || BC.

[blocks in formation]

Therefore, a str. line has been drawn through

A || BC.

Prop. 26.

Prop. 22.

Prop. 26.

PROP. XXXI. THEOR. 32. 1 Eu.

If a side of any triangle be produced, the ex-
terior angle is equal to the two interior and
opposite angles; and the three interior
angles of every triangle are together equal
to two right angles.

Let ABC be a A, and the side BC be
prod. to D. Then shall the ext. ▲ ACD =
CAB + ABC, the two int. and oppo.
Zs. And the 3 int. s, viz.,

BCA + ≤ CAB = 2 rt. ▲ s.

[blocks in formation]
[merged small][merged small][merged small][merged small][ocr errors]

Ax. 2

Ax. 2.

[blocks in formation]

Draw CE || AB;

AC meets them,

alt. BAC alt. / ACE.

:: BD falls upon the lines AB, CE,

[blocks in formation]


Z ABC, the int.

and oppo.

but LACE = BAC.

[ocr errors]

..whole ext. ACD = {BAC + <

to each of these equals add ACB,

[blocks in formation]




▲ ACD+≤ ACB = 2 rt. ▲ s. :: ≤ ACB + ▲ BAC+ / ABC = 2 rt. ▲ s. Wherefore, if a side of a triangle, &c.

COR. 1.-All the int. figure, together with 4 rt. twice as many rt. s as the

[blocks in formation]

Prop. 12. Ax. 1.

s of any rectilin.
s, are equal to
figure has sides.



For any rectilin. figure ABCDE can be divided into as many As as the figure has sides, by drawing str. lines from a pt. F within the figure to each of its angles. Then, by the preceding proposition, all the s of theses are equal to twice as many rt. s as there are As, i. e. as there are sides of the figure: and the same s are equal to the s of the figure, together with the s at the pt. F, which is the common vertex of the As; that is, together with 4 rt. s.

COR. 2.-All the ext. s of any rectilin. figure are together equal to 4 rt. s.

Cor. 2.
Prop. 14.

[blocks in formation]
« AnteriorContinuar »