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CASE II.-Let the vertex D of A ADB, be within the A ACB. C. ]| by Pst.1&2. Join C to D, & produce AC, AD,

to E and F. 1 byH.1&P.5 In A ACD, : AC = AD,

i. _ ECD

= _ FDC.
2 C.& Ax.9. But [ ECD is greater than

_ BCD;
.. FDC is greater than

_ BCD;
3 à fort. and much more is _ BDC > A

B _ BCD. 4 H. & P.5. Again in A BDC, ::: BD BC,.. _ BDC= _ BCD; 5 D. 3. But _ BDC is also > _ BCD; 6 D. 4, 3. :: BDC is both > and= _BCD, which is impossible. CASE III.- When the vertex of a A ADB is on the side of a À ABC, no demon

с stration is required. 7| Rec. Therefore, upon the same base &c. Q.E.I).

SCH.-The argument used in this proposition, is the Dilemma, or double Antecedent, in which the truth of the one is impossible, if we admit the truth of the other. The dilemma may, however, A have more than two antecedents; and Whately, p. 72, defines

B the true dilemma as “a conditional Syllogism, with several antecedents in the major, and a disjunctive in the minor.

PROP. 8.—THEOR.-(Important.) If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal; the angle which is contained by the two sides of the one, shall be equal to the angle contained by the two sides equal to them of the other.

DEM.-P. 7, Ax. 8.

D

2

B

E
E. 1 Нур. . Let the As ABC, DEF, have AB= DE, AC = DF, and

also BC · EF; Conc. then _ BAC shall equal / EDF. 1). 1 by Superp. Apply a ABC to A DEF, so that the · B be on · E,

and BC, on EF, 2 Hyp. .: BC

the · C coincides with the · F. 3 D.2, Hyp. Wherefore, :: BC coincides with EF, BA and CA shall

coincide with ED, FD. 4 Sup. 1. For, suppose that BC coincides with EF, 2. but not BA and CA with ED, FD, but with other lines,

as EG, FG,

EF,

D. 5 by Conc. then on this supp., in As EDF, EGF, on the same side

of EF, ED shall EG, and also FD = FG; 6 P. 7. which is impossible. 7 D. 2, 3. .-. since BC coincides with EF, the sides BA, CA, coin

cide with ED, FD: 8 D. 7. Wherefore, likewise _ BAC must coincide with _ EDF; 9 Ax. 8. and .. _ BAC is _EDF. 10 Rec. Therefore, if two triangles have two sides, &c. Q.F.D.

SCH.—The Equality estab'ished is that of the angles-hut the sides being equal, the triangles also must be equal. This is the second criterion of the equality of triangles.

APP.-By this Proposition, and Prop. 22, the angle at a given point C, made by lines from two objects, as A and B, may be determined

D

E without a theodolite.

A Measure the distances AB,BC, CA,- and from a scale construct a triangle DEF, the sides of which, DE, EF, and FD, will be representative of the distances AB, BC, and CA: then, with the semicircle find the number of degrees in angle F; and as the triangies ACB, DEF are similar, that number of degrees will also be the measure of angle C.

F

PROP. 9.-PROB.
To bisect a given rectilineal angle, i.e., to divide it into two equal parts.

SOL.-P.3, P. 1. & Pst. 1.-DEM.P. S.
E. l, Dat. Let the given rectil, _ be BAC;

2 Quæs. it is required to bisect it. C. 1 Assum. Take any point D in AB; 2 by P.3&Pst. 1 on AC make AE = AD,

and join DE, 3 P.1&Pst. 1 and on DE construct an equil. A, DFE, and join AF;

E E 4 Sol. then / BAF shall= _CAF,

the BAC being bisected. D. I by C. 2 & P. 1. In AS DAF, EAF,

.:: AD= AE, DF = EF,

and AF common, 2 ... DAF is = LEAF.

BI 3 Rec. Wherefore, the < BAC is bisected by AF.

Q.E.F. Sch.-1. The bisection of the arc DGE, which measures an angle, is also effected by the bisection of the angle.

2. By successive bisections an angle may be divided into any number of equal parts indicated by a power of two, as into four, eight, sixteen, thirty-two, &c. equal parts.

3. Hitherto no method has been discovered of trisecting an angle by Plane Geometry, so that the division of the quadrant of 90° into single degrees, is in part effected mechanically.

4. The Mariner's Compass is divided into its 32 parts, or points, by Props. 9 and 10. The bisection of the diameter by another diameter at rt. angles, gives the cardinal points N., E. S., W.; the quadrants, being bisected, give the points N.E., S.E., S.W., N.W.; and these bisections are continued until the number of equal divisions of the circle amounts to 32,the arc at each pair of points enclosing an angle of 119 15 minutes.

P.8.

2

PROP. 10-PROB.
To bisect a given finite st. line, i.e., to divide it into two equal parts

SOL.-P. 1, P. 9-DEM. P. 4.
E. 1, Dat. Given the st. line AB;

2 Quæs. it is required to bisect it.
C. 1 by P. l. On AB make an equil. A ABC;
P. 9. let CD make _ AČD = BCD,

and meet AB;
Sol. then AD = BD, i.e, ABis bisected

in D.
D. 1 by C. 1 & 2. In As ACD, BCD, ::: AC= BC,

CD is common and ACD =

_ BCD, P. 4. AD = DB. 3) Rec. Wherefore AB is bisected in D.

Q.E.F.

B

2

SCA.—By successive bisections a st. line may thus be divided into any number of equal parts indicated by a power of 2; as, 4, 8, 16, 32, 64, &c.

Prop. 11.-PROB. To draw a st. line at right angles to a given st. line from a given point in the same.

SOL.-P. 3, P. 1, Pst. 1.--DEM.-P. 8, Def. 10, Ax. 1. E. 1 Data Given the st. line

AB&Ca pointin it; 2 Quas. to draw from · C a st.

line at rt. angles to AB. C. 1 by P. 3. In AC take

any
D, & make CE

CD;
2 P.1&Pst.1. on DE construct

the equil, AFDE,
and join

FC;
3 Sol. then ZsDCF and
ECF are equal,& A D

C

E B consequently CF

is at rt, angles to AB. D. 1 by C.1&2. In As DCF, ECF, :;: DC = EC, FC is common,

and DF

EF.
P. 8. .. ZDCF Z ECF:
3

C. 2. And by construction they are adjacent angles;
Def. 10.1.-. the Zs DCF and ECF are rt. angles.
Rec. Wherefore from the given point, C, &c.

Q.E.F.
COR. - Hence it may be shown that two st. lines cannot have a common
segment.
E. 1) If possible, let the segment AB be common to the st. lines ABC

and ABD. C. 1 by P. 11. | At · B draw BE, at rt. Ls to AB.

2

B

4

D. 1 by Hyp. ... AB and BC make one st. line ABC,and

E LABE is a rt.angle; 2. Def, 10. ..LABE =_EBC: 3 Нур. .

Again, 'AB and BD

are supposed to form
one line ABD,and _

D
ABE is a rt. angle; A-

С Def. 10...LABE = EBD: 5 D.2,4,Ax.1.) Wherefore _ EBD= EBC; which is impossible. 6 Rec. Therefore two st. lines cannot have a common segment. Q.E.D.

Scu.-1. This corollary should be ranked among the axioms; for it is assumed in Prop. 4, that if certain lines placed on one another coincide for any portion of their length, they must coincide throughout. It is also assumed in Prop. 8.

2. When the given point is at the extremity of a given line, the line from that extremity should be produced, and the rt, angle be then constructed.

APP.--By this proposition the square is constructed, an instrument employed for ascertaining the perpendicular to a horizontal line, and for all purposes for which right angles are needed.

PROP. 12.-PROB. To draw a perpendicular to a given st. line of unlimited length from a given point without it.

Sol.-Pst. 3, P. 10, Pst. 1.--DEY.-Def. 15, P. 8, Def. 10.

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E. 1 Dat. Given the unlimited st. line AB, and the · C out of it;

2 Quæs. required from · C a perpendicular to AB. C. 1 Assum. Take any · D on the other side of AB; 2 by Pst. 3. from centre C, with rad. C]), draw the arc EGF, cut

ting AB in F and G; P.10, Pst. 1 bisect FG in H; join C and H, C and F, and C and G; 4 Sol. then CH is perpendicular to AB. D. 1 by C.3, Def.15 In As FHC, GHC, :;: FH = HG, CF=CG, and HC,

com. 2 P. 8.

.. L CHF = _ CHG: 3 C. Remk. Now these are adjacent angles; Def. 10.

CH is perpendicular to AB. 5 Rec. Therefore from the given point, &c.

Q.E.F. SCH.--1. The properties of the circle form the subject of the third book, but in the construction for the 12th Prop., the Lemma is borrowed from Prop. 2, bk. iii., that the circle will intersect the line in two points.

Use.—This problem is indispensable to all Artificers, Surveyors, and Engineers.

ABE,

PROP. 13.—THEOR. The angles which one st. line makes with another upon one side of it are either two right angles, or together equal to two right angles.

Con.-P. 11.-DEM.-Def. 10, Ax. 8, Ax. 2, Ax. 1. E. 1 Hyp. Let the st. line AB make angles with the st. line DC,

on one side of it, 2 Conc. 1. then the Zs CBA, ABD, are either two rt. angles, 3 2. or together = two rt. angles.

CASE I--Suppose that the ang. CBA is equal to ang. ABD; D. 11 by Def. 10. then each of the angles is a right angle.

CASE II.-But suppose that the angle CBA is not equal to angle ABD; C. 1 by P. 11. At · B in DC draw BE at rt. angles to CD,

2 Def. 10 then _s CBE, EBD are two ft. an gles. D. 1 by Ax. 8. | ::: L CBE,

= Ls CBA +

А
2 Add. add / EBD to each,
3 Ax. 2. .. Zs CBE, EBD, = Ls

CBA, ABE, and EBD.
Ax. 8. Again, :: _ DBA = Ls

DBE & EBA, 5 Add.

B
to both add _ ABC;
Ax. 2. .. Zs DBA + ABC= Ls
DBE, EBA and ABC.

E A
7 D. 2. But Zs CBE, EBD = Ls
DBE, EBA and ABC;

4

6

F

G 8 Ax. 1. .. Ls CBE, EBD = Ls

DBA, ABC; 9 C. Now Zs CBE, EBD are two

rt. angles, 10 Ax. 1. .. Zs DBA, ABC together D

= two rt. angles. 11 Rec. Wherefore, the angles which one st. line, g'c.

Q.E.D. 1. If one angle be a rt. angle, the other is a rt. angle; if one be obtuse, the other isacute: and if one be acute, the other is obtuse.

2. A semicircle is the measure of two right angles; and all the argles formed by any number of lines converging to one point, on one side of another line, are together equal to two right angles.

APP.—This theorem is of frequent use in Trigonometry and Astronomy. When we know one of the angles which a st. line meeting another st. line makes, at a point in it, we in fact know the other, for if we subtract the given angle from 180° we have the other angle: thus, let ang. ABC equal 700, ABD equals 180—70, or 110.

PROP. 14.-THEOR. (converse of the 13th.) If at a point in a st. line, two other lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two lines Pall be in one and the same st. line.

CON.-Pst. 2.--DEM.-P. 13. Ar. 1, As. 3.

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