(26.) Without the admission of the truth of a few first principles, which are called Axioms, no advance can be made in discovering the more intricate and complicated relations, which are unfolded in the subsequent Theorems. Besides admitting the truth of axioms, it is necessary to admit that certain very simple operations may be performed, such as are mentioned in the foregoing Postulates.

Any person who will admit the truth of these axioms, and grant the postulates, will be compelled to admit the truth of all subsequent theorems which shall be legitimately drawn from these first principles. And in this course of reasoning different individuals must arrive at precisely the same conclusions. It is for this reason that the study of Theoretical Geometry has always been regarded as one of the best methods of bringing out, and giving clearness and conciseness to the reasoning powers of the mind.

PROPOSITIONS.

PROPOSITION I.

THEOREM. When a straight line meets another straight line, the sum of the two adjacent angles, thus formed, is equal to two right-angles.

Let the straight line AB be met by the straight line CD at the point D. Then will the two adjacent angles ADC, BDC be together equal to two right-angles.

F

A

B

D

Suppose the line DF to be at right-angles to AB (Def. X.) The angle ADC is composed of the two angles ADF and FDC: therefore the sum of the two angles ADC, CDB is equal to the sum of the three angles ADF, FDC,

CDB, (Ax. II;) of which the first, ADF, is a right-angle, (Def. X,) and the sum of the other two, FDC and CDB, composes the right-angle FDB: therefore the sum of the two angles ADC and BDC is equal to two right-angles.

Cor. 1. Hence, also, conversely, if the two angles ADC, BDC, on the same side of the line AB, make up together two right-angles, then AD and DB will form a continued straight line AB.

Cor. 2. Hence, all the angles which can be made at any point D, by any number of lines on the same side of AB, are together equal to two right-angles.

Cor. 3. And as all the angles that can be made on the other side of AB are also equal to two right-angles; therefore all the angles that can be made around the point D, by any number of lines, are equal to four rightangles.

(27.) The instrument called a square, which is extensively used in the arts for tracing lines at right-angles to each other, consists of two flat rulers placed at rightangles, as in the adjoining figure. When much precision is required, great care should be taken by those purchasing this instrument, to test its accuracy. The preceding Proposition suggests a very simple and sure means of making such test.

Suppose the straight line AB to be the edge of a board, or any other plane surface. Apply one side of the square so as to coincide with AC; then, along the other edge, upon the surface, draw the line CD. Now, reversing the square, apply the first side so as to coincide

with BC; and then, along the second side, trace upon the surface the line CF. If the square is perfectly accurate, it is obvious that the lines CD CF will coincide.

This method not only detects an error in the instrument, when it exists, but also shows the amount of error; that is, how much the angle of the square exceeds or falls short of being a right-angle, or of containing just 900.

(28.) If a sheet of smooth, pliable paper be first folded, forming a straight edge at the fold, (Art. 7,) and afterwards it be again folded, so as to bring the folded edge upon itself, the angle at the folded corner will be very nearly a right-angle.

THEOREM.

PROPOSITION II.

When two lines intersect each other, the opposite angles are equal.

Let the two lines AB, CD intersect at the point F; then will the angle AFC be equal to BFD, and the angle AFD equal to BFC.

A

B

F

For, since the line CF meets the line AB, the two angles AFC, BFC, taken together, are equal to two rightangles (Prop. I.) In like manner, the line BF, meeting the line CD, makes the sum of the two angles BFC, BFD equal to two right-angles. Therefore the sum of the two angles AFC, BFC is equal to the sum of the two BFC, BFD (Ax. I.) And if the angle BFC, which is common, be taken away from each of these equals, the remaining angle AFC will be equal to the remaining angle BFD (Ax. III.) And in the same manner it may be shown that the angle AFD is equal to BFC.

PROPOSITION III.

THEOREM. If two triangles have two sides and the included angle of the one, equal to the two sides and the included angle of the other, the triangles will be identical, or equal in all respects.

In the two triangles ABC, DFG, if the side CA be equal to the side GD, and the side CB equal to the side GF, and the angle C equal to the angle G; then will the two triangles be identical, or equal in all respects.

For, conceive the triangle ABC to be placed upon the triangle DFG, in such a manner that the point C may coincide with the point G, and the side CA with the equal side GD. Then, since the angle G is equal to the angle C, the side CB will take the direction of the side GF. Also CB being equal to GF, the point B will coincide with the point F; consequently the side AB will coincide with DF. Therefore the two triangles are identical, and have all their other corresponding parts equal, (Ax. IX,) namely, the side AB equal to the side DF, the angle A equal to the angle D, and the angle B equal to the angle F.

Solution. Produce BD until DF is equal to BD: draw AF, cutting the mirror at G, and G will be the point required. For, comparing the triangle GFD with GBD, we have DF equal to DB, DG common, and the angle GDF equal to GDB, each being a rightangle; therefore those triangles are equal, (Prop. III :) consequently the angle DGB is equal to the angle DGF: but DGF is equal to the opposite angle AGC, (Prop. 11 :) hence the angle BGD is equal to AGC. Which is necessary, that the angle of incidence and the angle of reflection of the light may be equal. (30.) Suppose AC and BD to represent two trees standing on the horizontal plane AB, it is required to find a point in this plane equally distant from the tops C and D.

A G

B

Solution. Join CD, and bisect it by the perpendicular FG; then will the point G be the point sought. For, if we join GC, GD, we shall have the triangle GFC equal to GFD, since the side FC is equal to FD, the side FG common, and the angle CFG equal to DFG, each being a right-angle. Therefore (Prop. I,) the triangle GFC is equal to GFD; consequently GC is equal to GD.