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CASE II.-Let the vertex D of A ADB, be within the A ACB.
to E and F.
ni L ECD= LFDC.
| _ BCD;
_ BCD. 41 H. & P.5.1 Again in A BDC, :: BD = BC,:. / BDC= /BCD; 5 D. 3. But _ BDC is also > LBCD; 61 D. 4, 3. 1. 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
stration is required. 7| Rec. Therefore, upon the same base &c. Q.E.D.
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, have more than two antecedents; and Whately, p. 72, defines 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), 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: D. 7. Wherefore, likewise _ BAC must coincide with LEDF;
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
E without a theodolite.
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 triangles ACB, DEF are similar, that number of degrees will also be the measure of angle C.
SOL.-P.3, P. 1. & Pst. 1.-Dem.P. 8.
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. 11 and on DE construct an equil.
A, DFE, and join AF; 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 P. 8. ... / DAF is = Í EAF. B 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
e divided into any number of equal parts indicated by a power of two, as into four, eight, sixteen, thirty-two, &e. 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.
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.
D. I by Hyp. 1'.: AB and BC make
one st. line ABC,and
/ ABE is a rt.angle;
are supposed to form
ABE is a rt. angle; A4 Def. 10.12./ ABE = 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.
Sch.-1. This corollary should be ranked among the axioms; for it is assumed 1 Prop. 4, that if certain lines placed on one another coineide 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 might 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.
E. 11 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;
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. P. 8.
1. . _ CHF = L CHG: C. Remk. | Now these are adjacent angles; 4
Def. 10. 1.:. CH is perpendicular to AB. 51 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. ill., that the circle will intersect the line in two points.
USE.-This problem is indispensable to all Artificers, Surveyors, and Engineers.