A PROPOSITION is a distinct undivided portion of abstract science. It is either a problem or a theorem.

A PROBLEM proposes to effect some combination.

A THEOREM advances some truth, which is to be established.

A problem requires solution, a theorem wants demonstration; the former implies an operation, and the latter generally needs a previous construction.

A direct demonstration proceeds from the premises by a regular deduction.

An indirect demonstration attains its object, by showing that any other hypothesis than the one advanced involves a contradiction, or leads to an absurd conclusion.

A subordinate property, involved in a demonstration, is sometimes, for the sake of unity, detached, and then it forms a LEMMA.

A COROLLARY is an obvious consequence that results from a proposition.

A SCHOLIUM is an excursive remark on the nature and application of a train of reasoning.

The operations in Geometry suppose the drawing of straight lines and the description of circles, or they require in practice the use of the rule and compasses.

PROPOSITION I. PROBLEM.

To construct a triangle, of which the three sides are given.

Let AB represent the base, and G, H two sides of the triangle, which it is required to construct.

From the centre A with the distance G describe a circle, and from the centre B with the distance H describe another circle meeting the former

in the point C: ACB is the tri

angle required.

Because all the radii of the same circle are equal, AC is equal to G; and for the same

reason, BC is equal to H. Con

G

H

B

sequently the triangle ACB answers the conditions of the problem.

Corollary. If the radii G and H be equal to each other, the triangle will evidently be isosceles; and if those lines be likewise equal to the base AB, the triangle must be equilateral.

PROP. II. THEOREM.

Two triangles are equal, which have all the sides of the one equal to the corresponding sides of the other.

Let the two triangles ABC and DFE have the side AB equal to DF, AC to DE, and BC to FE: These triangles are equal.

Д

For let the triangle ACB be applied to DEF, in the same position. The point A being laid on D, and the side AC on DE, their other extremities C and E must coincide, since AC is equal to DE. And because AB is equal to DF, the point B must be found in the circumference of a circle described from D, with the distance DF; and for the same reason, B must also be found in the circumference of a circle described from E, with the distance EF: The vertex of the triangle ACB must, therefore, occur in a point which is common to both those circles, or in F the vertex of the triangle DFE. Consequently those two triangles, being rectilineal, must entirely coincide. The angle CAB is equal to EDF, ACB to DEF, and CBA to EFD; the equal angles being thus always opposite to the equal sides.

PROP. III. THEOR.

E A

Two triangles are equal, if two sides and the angle contained by these in the one be respectively equal to two sides and the contained angle in the other.

Let ABC and DEF be two triangles, of which the side AB is equal to DE, the side BC to EF, and the angle ABC contained by the former equal to DEF which is contained by the latter: These triangles are equal.

B

E

For let the triangle ABC be applied to DEF: The vertex B being placed on E, and the side BA on ED, the extremity A must fall upon D, since AB is equal to DE. And because the angle or divergence ABC is equal to DEF, and the side AB coin

CD

cides with DE, the other side BC must lie in the same direction with EF, and being of the same length, must en

tirely coincide with it; and consequently the points A and C resting on D and F, the straight lines AC and DF will also coincide. Wherefore, the one triangle being thus perfectly adapted to the other, a general equality must obtain between them: The third sides AC and DF are equal, and the angles BAC, BCA opposite to BC and BA are equal respectively to EDF and EFD, which the corresponding sides EF and ED subtend.

PROP. IV. PROB.

At a point in a straight line, to make an angle equal to a given angle.

At the point D in the given straight line DE to form an angle equal to the given angle BAC.

In the sides AB and AC of the given angle, assume the points G and H, join GH, from DE cut off DI equal to AG, and on DI constitute (I. 1.) a triangle DKI, having the sides DK and IK equal to AH and GH: EDF is the B/

angle required.

AA

For all the sides of the triangles GAH and IDK being respectively equal, the angles opposite to the equal sides must be likewise equal (I. 2.), and consequently IDK is equal to GAH.

Cor. If the segments AG, AH be taken equal, the construction will be rendered simpler and more commodious.

Scholium. By the successive application of this problem, an angle may be continually multiplied. Two circles CEG and ADF being described from the vertex B of the given angle with

C

radii BC and AB equal to its sides, and the base AC

being repeated between those circumferences; a multitude of triangles are thus formed, all of them equal to the original triangle ABC. Consequently the angle ABD is double of ABC, ABE triple, ABF quadruple, ABG quintuple, &c.

If the sides AB and BC of the given angle be supposed equal, only one circle will be required, a series of equal isosceles triangles being constituted about its centre. It is evident that this addition is without limit, and that the angle so produced may continue to swell, and its expanding side make repeated revolutions.

PROP. V. PROB.

To bisect a given angle.

H

Let ABC be an angle which it is required to bisect.

In the side AB take any point D, and from BC cut off BE equal to BD; join DE, on which construct the isosceles triangle DEF (I. 1.), and draw the straight line BF: The angle ABC is bisected by BF.

For the two triangles DBF and EBF, having the side DB equal to EB, the side DF to EF, and BF common to both, are (I. 2.) equal, and consequently the angle DBF is equal to EBF.

D

Cor. 1. It is evident that BG, the production of BF, di-vides the reversed angle ABC into two equal angles DBG and EBG.-The position of BG might also be determined, by the vertex of an isosceles triangle erected above DE and with sides greater than DB or EB.

Cor. 2. Hence the mode of drawing a perpendicularfrom a given point B in the straight line AC; for the an