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PROP. XXVIII. THEOR. 29. 1 Eu.

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.

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Prop. 12. but

AGH+ / BGH= 2 rt.

≤ s,

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s.

Ax. 12.

..

Hyp.

BGH+≤ GHD<2 rt.

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

AGH=/ GHD,

LAGH is not

Prop. 14.

i. e. but

< AGH=/ EGB,

Ax. 1.

[ocr errors]

GHD,

Ax. 2.

EGB=GHD;

add to each the BGH,

../EGB+ BGH=/ BGH+GHD,

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

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

Then GK cuts the lines AB, EF,
LAGH = ▲ GHF.

Prop. 28.

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

Prop. 28.

F;

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,

Z ADC,

then shall EF || BC.

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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.,
ABC +

BCA + ≤ CAB = 2 rt. ▲ s.

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Ax. 2

Ax. 2.

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Draw CE || AB;

AC meets them,

alt. BAC alt. / ACE.

:: BD falls upon the lines AB, CE,

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={

Z ABC, the int.

and oppo.

but LACE = BAC.

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..whole ext. ACD = {BAC + <

to each of these equals add ACB,

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LACB+2
BAC+/

ABC;

but

▲ 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

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Prop. 12. Ax. 1.

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

C

B

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.

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