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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.

GEN. ENUN.-To describe an isosceles ▲ on a given finite straight line.

PART. ENUN.-Let AB be the gn. st. line; then it is required to describe an isosc. A upon it.

CONST.-In AB, produced if neces

sary, take any pt. D, and make AE

=

F

Fig. 1.

G

Ꭰ E

B

ECF.

BD (Prop. III). From cr. A, with rad. AE, describe From cr. B, with rad. BD, describe DCG (Post. 3). From point of intersection c, join AC, BC (Post. 1). Then ABC is the ▲ required.

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 ▲ ABC is isosceles, and it has been described upon the gn. straight line AB.-Q. E. F.

F

Fig. 2.

C

DA

BE

PROP. IV. THEOREM.

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

[blocks in formation]

= DEF; another ACB shall = ▲ DFe. DEMONST.-1. For if the ABC be applied to the 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 DF, because the BAC EDF. (Hyp.) 3. Also, pt. c shall coincide with pt. F, because AC = DF. (Hyp.)

4. Since therefore the pt. в was proved to

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 ▲ ABC coincides with the area of the ▲ DEF, and is to it: and the others of the one coincide with the remaining of the other, and are = to them; viz. :

ABC DEF; and ACB = DFE. Therefore, if two As have two sides of the one to two sides of the other, 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 others, to which the equal sides are opposite.-Q. E. D.

S

This Proposition involves the first condition of the perfect equality of two ▲ in all their parts; and it extends of course to the equality of any number of ▲s under the same circumstances. The method of proof adopted in it is not by construction, but by supraposition: i.e. by laying one ▲ 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 A". The student will be assisted in comprehending the Proposition and its proof by forming two = 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.

PROP. V. THEOR.

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 = ACB, and ≤ CBD = ▲ BCE. CONST.-In BD take any pt. F.

D

F

A

B

C

G

E

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 mon to the two As AFC, AGB;

FAG, com... the base

FC = the base BG, and the area of ▲ AFC = area of ▲ AGB, and the remaining 4s of the one remaining s of the other, to which the equal sides are opposite (Prop. IV.): i.e. ▲ ACF = ABG, and AFC =

=

2. To prove that BF CG.

AGB.

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).

3. To prove ▲ BFC = A CGB.

Also it has been proved that FC = GB; in the BFC, CGB, the two sides BF, FC = the two sides CG, GB, each to each, and ▲ BFC = CGB; .. ▲ BFC = ACGB (Prop. IV.), and the remaining 4s are =, each to each: i. e. FBC ≤ GCB, and ▲ BCF =

4. To prove ABC = ≤ ACB.

CBG.

[blocks in formation]

ACB (Ax. 3): i. e. the

ABC remaining

=

Zs at the base of the ▲ ABC are = to one another.

5. And it has already been proved that the FBC GCB, which are the s on the

other side of the base.

Therefore the s at the base, &c.—Q. E. D. COROLLARY.-Hence every equilateral ▲ is

also equiangular.

A

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

C

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,

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