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

2 P. 3.

3 P. 31 & 3. 4 Conc.

5 Def. 35, C.

6 Conc.

D. 1 byC.3&2,P.29

2 P. 4.

3 Cor. P. 15.

Along EB produced set EB, as often as will make the distance EB, repeated, fall below CD; in the present instance EF equalling twice EB, or EB = BF;

From F draw FG || CD or AE, and = AE;

the line ABG is but one st. line.

And . CD cannot cut its || FG, and the line AB concurs with FG;

CD shall cut AG between B and G.

In As AEB, BFG, the side AE= FG, BE = BF, a and
LAEB = BFG;

/FBG:

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 D on the line BC, the angle CDA is found to be 60°: the distance from D to A is required.

From a scale construct the triangle ABC; and at B make the angle CBE equal to 60°; 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.

B

E

PROP. 32.-THEOR.-(Very Important.)

A

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.

D. 6 Add.

7 Ax. 2. 8 P. 13.

9 Ax. 1. 10 Rec.

COR. I. four rt. angles,

To each of these equals add the

ACB;

then /SACD+ACB= the threes BAC+ABC+ACB:
But s ACD + ACB = two rt. angles;

.. the threes BAC + ABC+AČB = two rt. S.
Wherefore, if a side of a triangle be produced, &c.

Q.E.D

All the interior angles of any rectilineal figure, together with 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

2.

2 Conc.

Fa within it;

,E; E

and let st. lines be drawn from F to
the angular points A, B, C, D,
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
sides,

2 P. 32.

and the three s in each ▲ = two rt. angles;

B

3 D. 1 & 2... all the angles of the triangles = two right angles 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 s at the F, the com. vertex of all the As,

.. all the int. s of the fig. + four rt. angles = two rt. angles the number of sides.

Q.E.D.

COR. II.-All the exterior angles of any rectilineal figure, as ABCDE, are together equal to four rt. angles.

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. 1 by Hyp.. the s ABC + ACB

2 P. 32.

=

Zs AEDADE;

and the angles ABC + ACB + BAC two rt. angles;

D. 3 P. 32.
4 Sub.
5 Ax. 3.

body.

also the angles AED + ADE + DAE= two rt. angles take away the equals ABC + ACB and AED + ADE; ..the rem. BAC the rem / DAE.

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 angle ZAS. For, by P. 32, the ext. angle ZAS is equal to the angles CS: 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.

Q.E.D.

B

B

D

E

At station A with a theodolite ascertain the angle CAD,-suppose 45°, measure AB 300 feet, and at station B measure the angle CBD, 309; 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 30°; and at A an angle CAD of 45°: 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.-DEм.-P. 29, P. 4, P. 27.

E. 1 Hyp. 1. Let the st. line AB be and to the

2 Conc.

st. line CD ;

2. and let the extremities towards the
same parts be joined, A to C, and B

to D ;

then the st. lines AC and BD are =

and .

C.

by Pst. 1. Join B and C.

D. 1 byH.&P.29. BC meets the Is AB, CD,

the ABC the alt.

=

BCD:

2 H. & D. 1. Hence,. AB = CD, BC com. to the two As, and ABC= ▲ BCD,

3 P. 4.

4 D. 3.

5 P. 27.

6 D. 3.

7 Rec.

4 ACB = CBD, and AC = BD.

And the st. line BC with the two st. lines AC, BD, makes angle ACB angle CBD,

..the st. line AC is | to the st. line BD;

And AC is equal to BD.

Wherefore, the st. lines which join the extremities, &c. Q.E.D.

APP.-This Theorem enables us to ascertain the perpendicular height of a mountain AG, as well as the distance from the base to the foot of the perpendicular, CG.

Take a large right-angled triangle, ABD and putting one end at A, let the side DB, by means of a plummet, be brought to the perpendicular; DB shows the altitude AH, and DA the horizontal distance BH: repeat the process as often as the case requires;-then DA + EB, &c., equals the horizontal distance CG; and DB + EC, &c., the altitude AG.

E

D

B

A

The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them; that, is, divides them into two equal parts.

4 P. 26.

5 D. 1 & 2.

6 Ax. 2.

7 D. 4.

8 Conc.

and thus BC bisects the .

C

BC meets the Is AB aud_CD,
the ABC the alt. BCD,
and ⚫. BC meets the Is AC and BD,
the ACB the alt. CBD :

B

Hence in As ABC, BCD, •.• Zs ABC, ACB = <s BCD and CBD, and BC is common;

.. ≤ BAC=< BDC, AB

CD, and ACBD.

And. ABC= ▲ BCD, and ▲ CBD = ≤ ACB, .. the whole ABD the whole

the opp. sides and angles of s are equal.

9 H. & D. 1 Also, . AB = CD, BC is common,

10 P. 4.

and ABC =/ BCD;

..the ▲ ABC the ABCD,

and the diameter bisects the

.

11 Rec. Wherefore, the opposite sides and angles of, &c.

SCH.-1. The two diagonals, AD and BC, bisect each other.

Q.E.D.

2. The converse to Prop. 34 is,-If the opposite sides oropposite angles of a quadrilatera

figure be equal, the opposite sides shall be parallel.

E.

Arr. 1-A finite st. line may be divided into any given number of equal parts.
Given a line AL and the

Dat,

3 P. 31. 4 Sol.

2P.3&Pst.1. take AB, and make BC,

A

EX

L

5 P. 31.

CD, DE each = AB, and join EL;

through D, C, B draw Dd, Cc, Bb, ||s to E,L;

the line AL is so divided that Ab=bc, bc=cd, and cd= dL.

Also through b draw b m ||

D. 1byC3&5P34. Bb Cm is a,.. bm

2 P. 29.

3 P. 26.

4 Sim.

also A

=

AX.

BC or AB;

cbm, and

AbB = ≤ b cm:

Q.E.D.

therefore Ab is equal to bc.

And in the same way dc-bc, and dL=cd.

2. On the same principle the Sliding Scale, called from the inventors the Vernier or Nonius, is constructed. This scale is very useful, as in the Barometer.

For measuring the hundreth part of an inch; we may take Ritchie's description, in his Geometry, p. 32. 'Suppose an inch AB, divided into ten equal parts, each part will be

1-10th of an inch. 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.