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from AB, the > of two gn. st. lines, a part AE has been cut off, which is = to c, the < .Q. E. F.

Of this Proposition there are also several cases, inasmuch as the given lines need not be separated ; but the shorter may be attached either to the extremity of the other, when the application of Prop. II. becomes unnecessary; or the two lines may meet or intersect each other somewhere between the extreme points ; or, again, the less may be a part of the greater. The student will do well to prove each case, but it is superfluous to insert them here. We are now, however, in a condition to give the following general solution of Prop. A. The principle is carried yet further in Prop. XXII.

PROP. B. PROB.

G

D E

B

GEN. ENUN.- To describe an isosceles A on a given finite straight line. PART. ENUN.—Let AB be the gn.

Fig. 1. st. line; then it is required to describe an isosc. A upon it. CONST.-In AB, produced if neces

A sary, take any pt. D, and make AE BD (Prop. III). From cr. A, with rad. AE, describe O ECF. From cr. B, with rad. BD, describe O DCG (Post. 3). From point of intersection c, join AC, BC Fig. 2. (Post. 1). Then ABC is the A required. F

DEMONST.-By the property of the O, AE = AC, and BD = BC (Def. 15). But, by construction AE =

BD; .. also AC = BC (Ax. 1). Hence the A ABC is isosceles, and it has been described upon the gn. straight line AB.-Q. E. F.

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D A

BE

PROP. IV. THEOREM.

B

с

GEN. ENUN.-If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal, they shall likewise have their bases, or third sides, equal; and the two triangles shall be equal; and their other angles shall be equal, each to each; viz. those to which the equal sides are opposite.

PART. ENUN.-Let ABC, DEF, be two As, which have the two sides AB, AC = to the two sides DE, DF, each to each; viz. AB = DE, and AC = DF,

and also the < BAC = Z EDF; then the base BC shall base EF;

and the area of the A ABC shall area of a DEF; also the < ABC shall = 2 DEF; another < ACB shall =

Z DFE. DEMONST.-1. For if the A ABC be applied to the A DEF, so that the pt. A may be on D, and the st. line AB upon DE; the pt. B shall coincide with the pt. E, because AB = DE. (Hyp.)

2. Also, AB coinciding with De, Ac shall fall upon Dr, because the BAC = _ EDF. (Hyp.)

3. Also, pt. c shall coincide with pt. F, because ac = DF. (Hyp.)

4. Since therefore the pt. B was proved to

=

E

coincide with the pt. E, the base bc must coincide with the base EF; because, the pts. B and c coinciding respectively with E and F, if the base bc does not coincide with the base EF, but lies above or below it, two straight lines would inclose a space, which is impossible (Ax. 10); .:. the base bc coincides with the base EF, and is = to it (Ax. 8).

5. ... the whole area of the A ABC coincides with the area of the A DEF, and is = to it: and the other Zs of the one coincide with the remaining of the other, and are = to them; viz. :

2 ABC = Z DEF; and 2 ACB = 2 DFE.

Therefore, if two As have two sides of the one = to two sides of the ot

each to each, and have likewise the < $ contained by those sides equal, their bases shall likewise be equal ; and their areas shall be equal; and also their other <$, to which the equal sides are opposite.—Q. E. D.

This Proposition involves the first condition of the per. fect equality of two As in all their parts; and it extends of course to the equality of any number of AS under the same circumstances. The method of proof adopted in it is not by construction, but by supraposition: i. e. by laying one Å upon the other, and shewing that they coincide. It is again employed in the Eighth Proposition, which, together with the present, is the foundation of all that follows with respect to the comparison of As. The student will be assisted in comprehending the Proposition and its proof by forming

As of card-board, and placing one upon the other as directed in the text.

Upon this, and its cognate propositions, is based the Theory of Perspective.

two =

PROP. V. THEOR.

A

B

с

F

E

Gen. Enun.The angles at the base of an isosceles triangle are equal; and if the equal sides be produced, the angles upon the other side of the base shall be equal.

PART. ENUN.-Let ABC be an isosc. A, of which the side AB = side ac (Def. 24); and let the st. lines AB, AC be produced to D and E (Post. 2); then < ABC = ZACB, and Z CBD = ZBCE. D Const.—In BD take any pt. F.

From AE the > cut off ag = AF, the < (Prop. III)., and join FC, GB (Post. 1).

DEMONST.–1. To prove the As AGB, AFC = in every respect. Because AF = AG, and AB

= AC; ... the two sides FA, AC = two sides Ga, AB,

each to each. Also they contain — FAG, common to the two AS AFC, AGB; .:. the base FC = the base BG, and the area of A AFC = area of A AGB, and the remaining of the one = remaining 2$ of the other, to which the equal sides are opposite (Prop. IV.): i.e. Z

2 ABG, and Z AFC = Z AGB.
2. To
prove

that BF = CG. Again : because the whole AF = the whole AG (Const.), and the part AB = the part ac (Hyp.); .. the remainder BF = the remainder CG (Ax. 3).

ACF =

A CGB.

Z CGB ;

A BFC

3. To prove A BFC =

Also it has been proved that FC = GB; in the A$ BFC, CGB, the two sides BF, FC = the two sides cG, GB, each to each, and < BFC =

A CGB (Prop. IV.), and the remaining zs are =, each to each : i. e. _ FBC = Z GCB, and Z BCF = Z CBG.

4. To prove 2 ABC = < ACB.

Now since it has been shown that the whole L ABG = the whole Z ACF, and that the part < CBG = the part < BCF; .. the remaining Z ABC = remaining ACB (Ax. 3): i.e. the zs at the base of the A ABC are = to one another.

5. And it has already been proved that the Z FBC = Z GCB, which are the zs on the other side of the base.

Therefore the _s at the base, &c.—Q. E. D.

COROLLARY.—Hence every equilateral o is also equiangular.

The proof of this Corollary is very simple. Thus :Let ABC be an equilateral A; then, because A AB = AC, .. _ ABC = Z ACB; and because AB = BC, .. 2 BAC = L ACB; ... the 28 ABC, ACB, BAC, are equal : i.e. the A ABC is equiangular.-Q. E. D.

Partly from the figure, and partly from B the difficulty which the demonstration presents to the beginner, the Fifth Proposition has obtained the ominous title of Pons Asinorum, or the Asses' Bridge; and it certainly requires some attention to surmount it. The former part of it may be very readily proved by the method of supraposition ; and when Prop. XIII, has been proved, a still more simple demonstration of the whole will be given. In the meantime,

C

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