Q. But is it not possible for this line AC, to fall on the other side of the perpendicular?

A. No. Because the line AC, being greater than the line AB, would in this case be farther from the perpendicular, than the line AB (conseq. 4, preceding Query), and the angle at B would then fall without the triangle and because the whole triangle ABC is entirely determined, when two of its sides, and the angle, which is opposite to the greater of them, are given: therefore, all triangles, in which these three things are equal, must be equal to one another.

Q. What truth can you infer from this, respecting the case where the hypothenuse,* and one side of a right-angular triangle, are equal to the hypothenuse and one of the sides in another right-angular triangle?

A. That these two right-angular triangles must be equal to each other. For, in this case, we have two sides, and the right angle which is opposite to the greater of them in the one, equal to two sides, and the angle which is opposite to the greater of them, in the other.

*In a right-angular triangle, the side, which is opposite to the right angle, is called hypothenuse.

Q. But if in Fig. II. (page 54) the two sides AC, AB, and the angle at C, opposite to the smaller side AB, be given, would not this be sufficient to determine the triangle ABC?

A. No. For the two lines, AB, AE, being equal, there would be two triangles, ABC and AEC, possible, containing the same three things, and it would be doubtful which of the two triangles was meant.

QUERY XI.

If you have two sides, ab, bc, of a triangle abc, equal to two sides AB, BC, of another triangle ABC, each to each; but the angle ABC included by the two sides, AB, BC, in the triangle ABC, greater than the angle abc, included by the sides, ab, bc, in the triangle a b c, what remark can you make with regard to the two sides, a c, AC, which are respectively opposite to those angles?

A. That the side a c, opposite to the smaller angle a b c, in the triangle a b c, is smaller than the side AC, opposite to the greater angle ABC, in the triangle ABC.

Q. How do you prove this?

A. By placing the triangle abc upon the triangle ABC, with the side ab upon AB (its equal), the side b c will fall within the angle

ABC, because the angle a b c is smaller than the angle ABC; and the extremity c, of the line bc, will either fall without the triangle ABC, as you see in the figure before you, or within it, or it may also fall upon the line AC itself.

1st. If it falls without the triangle ABC, by imagining the line Cc drawn, the triangle c BC will be isosceles; for we have supposed the sides, bc, BC, to be equal; therefore, 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 angle y alone; and if the angle zis 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 AC c, the angle A c C is greater than the angle AC c; consequently the side AC, opposite to the greater angle Ac C, must be greater than the side a c, opposite to the smaller angle AC c.

2dly, If the extremity of the line bc falls within the triangle ABC, the sum of the two sides, a c, bc, must be small

a

er than the sum of the two sides, AC, BC (Query 8, A Sect. II.); therefore, by taking from each of these sums

a

b

the equal lines, bc, BC, respectively, the remainder AC, of the greater sum (AC+ BC),

will be greater than the remainder a c, of the smaller sum (a c + b c).

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 A c.

will the two triangles, ABC, CDB, bear to each other?

A. They will be equal to one another, and the parallelogram will be 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 11, Sect. I.) therefore, because the triangle ABC has a side CB, and the two adjacent angles, x and

w, equal to the side CB, and the two adjacent angles, y and z, in the triangle CDB, each to each, these two triangles must be equal (Query 6, Sect. I.), and the diagonal CB, must divide 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