the line Cc drawn, the triangle cBC will be isosceles; for we have supposed the side bc equal to BC; and because the angles at the basis of an isosceles triangle are equal (Query 3, Sect. II.), the angle z is equal to the sum of the two angles x and y; consequently greater than the angley alone; and if the angle z is greater than the angle y, the two angles z and w together will be greater still than the same angle y; therefore, in the triangle ACc, the angle AcC is greater than the angle ACc; conscquently the side AC, opposite to the greater angle AcC, must be greater than the side ac, opposite to the smaller angle ACc.

2dly. If the extremity of the line bc falls within the triangle ABC, the sum of the two sides ac, bc, must be smaller than the sum of the two

sides AC, BC (Query 8, Sect. II.); therefore, by taking from each of A these sums the equal lines bc, BC, respectively, the remainder, AC, of a

b

the greater sum (AC+BC) is greater than the remainder, ac, of the smaller sum (ac+bc).

Finally. If the point c falls upon the line AC itself, it is evident that the whole line AC must be greater than its part Ac.

A. They are equal to each other, and the parallelogram is divided into two equal parts.

Q. How can you prove this?

A. The two triangles, ABC and CDB, have the side CB common; and the angle y is equal to the angle w, because y and w are alternate angles, formed by the intersection of the two parallel lines CD, AB, by a third line, CB; and the angle x is equal to the angle z, because these two angles are formed in a similar manner, by the parallel lines AC, DB (Query 10, Sect. I.): and as the triangle ABC has a side CB, and the two adjacent angles, x and w, equal to the same side CB, and the two adjacent angles, and y, in the triangle CDB, each to each; therefore these two triangles are equal (Query 6, Sect. I.), and the diagonal CB divides the parallelogram into two equal parts.

Q. What other properties of a parallelogram can you infer from the one just learned?

1st. The opposite sides of a parallelogram are equal; that is, the side CD is equal to the side AB, and the side CA to the side DB; for in the equal triangles, ABC, CDB, the equal sides must be opposite to the equal angles. (Conseq. of Query 1, Sect. II.)

2dly. The opposite angles in a parallelogram are equal; for in the two equal triangles, ABC, CDB, the same side, CB, is opposite to each of the angles, at D and A. (Conseq. of Query 6, Sect. I.)

3dly. By one angle of a parallelogram, all four are determined; for the sum of the four angles in a parallelogram is equal to four right angles; because the sum of the three angles in each of the two triangles, ABC, CDB, is equal to two right angles. Now, if the angle at D, for instance, is known, the angle at A is equal to it; and there remain but the two angles ACD and ABD, each

of which must be equal to half of what is wanting to complete the sum of the four right angles.

Q. If you have a quadrilateral, in which the opposite sides are respectively equal, does it follow that the figure must be a parallelogram?

A. Yes. For if, in the last figure, you have the side CD equal to the side AB, and the side AC equal to the side BD; by drawing the diagonal BC, you have the three sides of the triangle ABC, respectively, equal to the three sides of the triangle CDB; therefore, these two triangles are equal; and the angle y, opposite to the side DB, is equal to the angle w, opposite to the equal side AC; and the angle x, opposite to the side AB, is equal to the angle z, opposite to the equal side CD; that is, the alternate angles, y and w, x and z, are respectively equal : therefore the side CD is parallel to the side AB, and the side AC to the side BD, and the figure is a parallelogram.

Q. If, in a quadrilateral, you know but two sides to be equal and parallel, what will then be the name of the figure?

A. It will still be a parallelogram. For if, in the last figure, the side CD is equal and parallel to AB, by drawing the diagonal CB, you have the two sides, CB and CD, in the triangle CDB, equal to the two sides, CB, AB, in the triangle ABC, each to each; and because the side CD is parallel to the side AB, the included angle y is equal to the included angle w; therefore the two triangles are equal (Query 1, Sect. II.), and the side AC is also equal and parallel to the side DB, as before.

that if, from the vertex A, for instance, you draw the diagonals AF, AE, AD, AC, to the vertices F, E, D, C, each of the two triangles AGF, ABC, will need for its formation two sides of the figure, and a diagonal; but then every one remaining side of the figure will, together with two diagonals, form a triangle; therefore there will be as many triangles formed, as there are sides less the two, which are additionally employed in the formation of the two triangles AGF, ABC.

Q. And what is the sum of all the angles, BAG, AGF, GFE, FED, EDC, DCB, CBA, equal to ?

A. To as many times two right angles as the figure ABCDEFG has sides less two. For as every rectilinear figure can be divided into as many triangles as there are sides less two; and because the sum of the three angles in each triangle is equal to two right angles (Query 13, Sect. I.) there will be as many times two right angles in all the angles of your figure, as there are triangles; that is, as many as the figure has sides less two.

SECTION II.

PART II.

OF GEOMETRICAL PROPORTIONS,* AND SIMILARITY OF TRIANGLES.

WHENEVER We compare two things with regard to their magnitude, and inquire how many times one is greater than the other, we determine the ratio which these two things bear to each other. If, in this way, we find out that the one is two, three, four, &c. times greater than the other, we say that these things are in the ratio of one to two, to three, to four, &c.: e. g. If you compare the fortunes of two persons, one of whom is worth $10,000, and the other $20,000, you say, that their fortunes are in the ratio of one to two. Or if you compare two lines, one of which is two, and the other six feet long, you say of these lines, that they are in the

* It is the design of the author to give here a perfectly elementary theory of geometrical proportions, and to establish every principle geometrically, and by simple induction. Intending the above theory for those who have not yet acquired the least knowledge of Algebra, he is not allowed to identify the theory of proportions with that of algebraic equations (as it is done by some writers on Mathematics), and then to find out the principles of the former by an analysis of the latter. There are several disadvantages inseparable from the algebraic method of considering a ratio as a fraction, besides the difficulty of making such a theory accessible to beginners. Neither can an algebraic demonstration be made obvious to the eye like a geometrical one.