In all reasoning definitions are necessary, in order to insure in the minds of the proponent and respondent identity of ideas. A corollary is an inference deduced from a previous course of reasoning. An axiom is a proposition evident at first sight. In the following demonstrations there are many axioms taken for granted (such as, things equal to the same thing are equal to one another, etc.); these it was thought not necessary to introduce in form.
If a straight line, as A B (Fig. 249), stand upon another straight line, as CD, so that the two angles made at the point B are equal - A B C to A BD (Art. 499, obtuse angle) - then each of the two angles is called a right angle.
The circumference of every circle is supposed to be divided into 360 equal parts, called degrees; hence a semicircle contains 180 degrees, a quadrant 00. etc.
The measure of an angle is the number of degrees contained between its two sides, using the angular point as a centre upon which to describe the arc. Thus the arc C E (Fig. 250) is the measure of the angle C B E, E A of the angle EB A, and A D of the angle ABB.
As the two angles at B (Fig. 249) are right angles, and as the semicircle, CAD, contains 180 degrees ((Art. 327), the measure of two right angles, therefore, is
180 degrees; of one right angle, 90 degrees; of half a right angle, 45; of one third of a right angle, 30, etc.
In measuring an angle (Art. 328), no regard is to be had to the length of its sides, but only to the degree of their inclination. Hence equal angles are such as have the same degree of inclination, without regard to the length of their sides.
If two straight lines parallel to one another, as A B and CD (Fig. 251), stand upon another straight line, as E F, the angles A BF and CDF are equal, and the angle A B E is equal to the angle C D E.
If a straight line, as A B (Fig. 250), stand obliquely upon another straight line, as CD, then one of the angles, as A B C, is called an obtuse angle, and the other, as A B D, an acute angle.
The two angles A BD and ABC (Fig. 250) are together equal to two right angles (Arts. 326, 329); also, the three angles A BD, EB A, and CBE are together equal to two right angles.
Hence all the angles that can be made upon one side of a line, meeting in a point in that line, are together equal to two right angles.
Hence all the angles that can be made on both sides of a line, at a point in that line, or all the angles that can be made about a point, are together equal to four right angles.
If to each of two equal angles a third angle be added, their sums will be equal. Let ABC and D E F (Fig. 252) be equal angles, and the angle IJ K the one to be added. Make the angles G B A and HE D each equal to the given angle IJ K; then the angle G B C will be equal to the angle HE F; for if A B C and D E F be angles of 90 degrees, and IJK 30, then the angles GBC and HEF will be each equal to 90 and 30 added, viz., 120 degrees.
Triangles that have two of their sides and the angle contained between them respectively equal, have also their third sides and the two remaining angles equal; and consequently one triangle will every way equal the other. Let ABC (Fig. 253) and D E F be two given triangles, having the angle at A equal to the angle at D, the side A B equal to the side D E, and the side A C equal to the side D F; then the third side of one, B C, is equal to the third side of the other, E F; the angle at B is equal to the angle at E; and the angle at C is equal to the angle at F. For if one triangle be applied to the other, the three points B, A, C, coinciding with the three points E, D, F, the line BC must coincide with the line E F; the angle at B with the angle at E; the angle at C with the angle at F; and the triangle B A C be every way equal to the triangle EDF.
The two angles at the base of an isosceles triangle are equal. Let ABC (Fig. 254) be an isosceles triangle, of which the sides, A B and A C, are equal. Bisect the angle (Art. 506) B A C by the line A D. Then, the line B A being equal to the line A C, the line A D of the triangle E being equal to the line A D of the triangle F (being common to each), the angle BAD being equal to the angle DA C, - the line BD must, according to Art. 337, be equal to the line D C, and the angle at B must be equal to the angle at C.
A diagonal crossing a parallelogram divides it into two equal triangles. Let CD E F (Fig. 255) be a given parallelogram, and CF a line crossing it diagonally. Then, as E C is equal to FD, and EF to C D, the angle at E to the angle at D, the triangle A must, according to Art. 337, be equal to the triangle B.
Let J KLM (Fig. 256) be a given parallelogram, and KL a diagonal. At any distance between J K and LM d raw NP parallel to J K; through the point G, the intersection of the lines KL and N P, draw HI parallel to KM. In every parallelogram thus divided, the parallelogram A is equal to the parallelogram B. For, according to Art. 339, the triangle JKL is equal to the triangle KLM, the triangle C to the triangle D, and E to F; this being the case, take D and F from the triangle KLM, and C and E from the triangle J K L, and what remains in one must be equal to what remains in the other; therefore, the parallelogram A is equal to the parallelogram B.