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C. 1 by Assum. P31 In AB take any · B, and through B draw EF || to AC; 2 P. 3.

Along EB produced set EB, as often as will make the distance EB, repeated, tall below CD ; in the present

instance EF equalling twice EB, or EB = BF; 3 P. 31 & 3. From F draw FG || CD or AE, and = AE; 4 Conc.

the line ABG is but one st. line. 5 Def. 35, C. And :: CD cannot cut its || FG, and the line AB

concurs with FG; 6 Conc. . . CD shall cut AG between B and G. D. 1 byC.3&2,P.29 In As AEB, BFG, the side AE=FG, BE =BF, and

- AEB = LBFG; 3 P. 4.

therefore the Z EBA = / FBG: 3 Cor. P. 15. consequently AB and BG make one st. line, and CD

being produced cuts AG in the point H. APP.--1. The drawing of parallel lines is required in Perspective, Navigation, the Construction of the Sector, or Compass of Proportion, and in other branches of practical Mathematics.

2. The use of parallel lines enables the Surveyor to ascertain the distance of an inaccessible object, by the method of Representative Values or of Construction : thus, there are three objects, A, B, C, distant from each other AC 6 miles, AB 8, and BC 9.4 miles ; at Don the line BC, the angle CDA is found to be 609: the distance from D to A is required.

From a scale construct the triangle ABC; and at B make the angle CBE equal to 609; AD, drawn parallel to BE, is the representative value of the distance from D to A. Taking DA in the compasses, and applying the distance to the scale, the measurement of DA will be found equal to 6 miles.

Prop. 32.—THEOR.—(Very Important.) If a side of any triangle be produced, the ext. angle is equal to the two interior and opposite angles ; and the three int, angles of every triangle are together equal to two rt. angles.

Con.-P. 31.-DEM.-P. 29, Ax. 2, P. 13, Ax. 1. E. 1. Hyp. | Let the side BC of the A

ABC be produced to D; 2 Conc. 1. | then the ext. 2 ACD =

the two int. and opp. Ls

CAB+ ABC;
, 2. ) and the three int. / s ABC

+ ACB + CAB = two

rt. angles. C. by P. 31. Draw from . C a st. line

1 CE || the side BA. D. 1byC.&P.29. .: AB is || to CE, and AC meets them,

.. the Ż BAC = the alt. LACE:

Again, . AB is to CE, and BD falls on them, 3 P. 29 1. the ext. L ECD = the int. and opp. L ABC: 4 D. 1. But ACE = /BAC; 5| Ax. 2. 1.. the whole ext. L ACD = the two int. and opp. Ls,

I BAC and ABC:

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D. 6 Add. To each of these equals add the L ACB ;

71 Ax. 2. then _s ACD+ ACB= the three ZsBAC+ ABC+ACB: 8 P. 13. But Zs ACD + ACB = two rt. angles ; 9 Ax. 1. 1.:. the three s BAC + ABC + AČB = two rt. Zs. 10| Rec. Wherefore, if a side of a triangle be produced, &c. Q.E.D)

COR. I.- All the interior angles of any rectilineal figure, together with four rt. angles, are equal to twice as many rt. angles as the figure has sides.

DEM.—P. 32, P 15, Cor. 2, Ax. 1, Ax. 2, Ax. 3.
E. 1 Hyp. 1. | Let ABCDE be a rectil. figure, and

Fa · within it;
, 2. and let st. lines be drawn from F to

the angular points A, B, C, D), E; R: 2 Conc. then all the int. Ls together with

four rt. angles = twice as many

rt. angles as the figure has sides. D. 1) by Hyp. ... then the fig. ABCDE is divided

into as many triangles as there are A

sides, 2 P. 32. and ... the three Zs in each A = two rt. angles ; 3 D. 1 & 2. ... all the angles of the triangles = two right angles X the

number of sides; 4 Hyp. But the same _s of these As = the int. angles of the

fig. + the angles at · F; 5 P15,Cor2 and all the Zs at the · F, the com, vertex of all the As,=

I four rt. angles ; 6 Conc. 1.:. all the int. Ls of the fig. + four rt. angles = two rt.

angles X the number of sides. COR. II. - All the exterior angles of any rectilineal figure, as ABCDE, are together equal to four rt. angles. C. | by Pst. 2 Produce all the sides of the

fig. ABCDE. D. 1 by P. 131: each int. 2 + the adj.

ext. L = two rt. angles ; 2 P. 32. 1.:. all the int. Zs + all the Cor. 1. adjacent ext. Zs=two rt.

A ----Zs x the number of sides. 3 Ax. 1. J and ... these = all the int.

| Ls + four rt. angles. 4 „ 2. 1., all the ext. Zs together F--------

T = four rt. angles. Q.E.D. CoR. III.-If two triangles ABC, ADE have two angles of the one, ABC, ACB, respectively equal to two angles of the other, AED, ADE; then the third angle of the one, BAC, shall be equal to the third angle of the other, DAE. D. 11 by Hyp... the Zs ABC + ACB = 28 AED + ADE : - P 32, \ and the angles ABC + ACB + BAC = two rt. angles ;

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D. 3 P. 32. I also the angles AED + ADE + DAE = two rt. angles

4 Sub. take away the equals ABC + ACB and AED + ADE ; 5 Ax. 3. l:. the rem. _ BAC = the rem (. DAE. Q.E.D. APP.-1. In Astronomy to determine the Parallax of a heavenly

Let C represent the earth's centre, A a point on the earth's circumference AB, and Z the zenith of the station A; S is a star or any heavenly body not in the zenith.

By observation take the angle ZAS, the zenith distance at the earth's surface A: if the star S were viewed from C the centre, the angle ZCS would be its zenith distance; and the angle ZCS is less than anglo ZAS. For, by P. 32, the ext. angle ZAS is equal to the angles C +S: thus, the ang. S is equal to the excess of the angle ZAS above the angle Zcs. If now, from an Astronomical Table, I learn what is the zenith distance of S as viewed from C, the earth's centre, the difference between ang. ZAS and ang. ZCS, or the angle ASC will be the parallax.

3. To construct a figure which will give the representative value of the perpendicular height of a mountain as CD.

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At station A with a theodolite ascertain the angle CAD,-suppose 45°, measure AB 300 feet, and at station B measure the angle CBD, 30°; and from these data construct the figure required.

From B draw a line of indefinite length BE, and take on it 300 equal parts to A: at B draw an angle CBD of 300; and at A an angle CAD of 450 : the lines BC and AC intersect in the point C; and a perpendicular from C to BE, namely CD, will represent the perpendicular height of the mountain, Take CD in the compasses, and apply the distance to the same scale; it will be found that CD measures 400 feet or yards, according to the unit of length in AB.

PROP. 33.-THEOR. The st. lines which join the extremities of two equal and parallel st. lines towards the same parts, are also equal and parallel.

Con.-Pst. 1.-DEM.-P. 29, P. 4, P. 27.
E. 1; Hyp. 1. , Let the st. line AB be= and || to the

st, line CD ;
, 2.and let the extremities towards the

same parts be joined, A to C, and B

to D; 2 Conc. then the st. lines AC and BD are =

and Il.

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nch. Again, suppose a line CD equal to 9 of these parts to be also divided into 10 equal parts, each of these will be 1-10th of 9-10th of an inch, which is 9-100th. But the length of one of the first divisions being 1-10th, or 10-100th, and that of the second 9-100th one of the first divisions is 1-100th of an inch longer than one of the second. If the line CD slide along parallel to AB till the two divisions marked 1, 1, form a continuous line, the sliding scale will have moved 1-100th part of an inch towards B. If it slide along till the next two divisions coincide, it will have moved 2-100th of an inch, &c.

PROP. 35.-THEOR. : Parallelograms upon the same base and between the same parallels are equal, or rather equivalent to one another.

DEM:-P. 34, Ax. 6, Ax. 1, Ax. 3, P. 29, P. 24.

CASE I. Sup. Let the sides AD, DE opp. to

BC be terminated in the
same point D.

BC
E. 1 Hyp. 1. | Let the O s ABCD, DBCE, be on the same base BC,

, 2. ) and between the same s AE and BC;
2 Conc. then the ABCD = the O DBCE.

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