6 MATHEMATICS. 
the greater side, and vice versa. If, in the triangle abc, (fig. 14), ab 1s 
greater than ac, then is also 7 ach greater than / abc, &c. 3. If two 
triangles have two sides of the one equal to two sides of the other, each to 
each, and the included angle unequal, the third sides will be unequal, and 
the greater side will belong to the greater triangle, which has the greater 
included angle ; or if an angle of a triangle is increased, while its including 
sides remain the same, then will the side opposite the angle be also increased 
( figs. 15-17). 
A quadrilateral is a parallelogram, when 1t can be proved,—1, that two of 
its opposite sides are equal and parallel; or 2, that all the opposite sides 
are equal; or 3, that all the opposite sides are parallel. The opposite angles 
of a parallelogram are also equal: two adjacent angles form together two 
right angles (fig. 22). 
Hence all the angles in a parallelogram are either right angles or two 
are acute and two obtuse (rhombus and rhomboid). The two diagonals of 
a parallelogram mutually bisect each other (fig. 19). 
To construct a parallelogram it is generally necessary to have two sides 
and an angle given. Fora rectangle, however, two sides are sufficient ; for 
a rhombus, one side and an angle; and for a square, one side. A trapezoid 
can be constructed by means of its four sides (pl. 3, fig. 21), by construct- 
ing first a triangle out of the two sides ad, ae, which are not parallel, and 
the difference of the parallel sides de, then producing de to c, so that ce may 
become equal to the lesser parallel, and with aec, the parallelogram abce 
will be completed. A trapezium may be constructed,—1, with four sides 
and an angle; 2, with three sides and the two included angles; 3, with 
three sides and the angles lying about the unknown side; 4, with two 
adjacent or two opposite sides and the three angles (figs. 23, 24). 
Every regular polygon has a central point which is at an equal distance 
from all its sides and from all its angles. This point is found by bisecting 
two angles, for the bisecting lines always meet in that common centre 
( fig. 25). 
The preceding propositions concerning the equality of triangles, enable 
us to solve and prove the solution of a number of problems of easier 
construction. Among these are: 1. to construct a triangle equal to a 
given triangle (jig. 26). 2. To describe a given angle m, on a given 
straight line ab, at a given point a (fig. 27). 3. Through a given point, to 
draw a line parallel to a given line. 4. To bisect a given angle (pl. 2, fig. 
77). 5. To bisect a given line (fig.75). 6. To draw a perpendicular to a 
line at a given point. 7. From a given point out of a line, to let fall a 
perpendicular on that line (pl. 3, fig. 28, and pl. 2, fig. 75). 8. To draw a 
perpendicular at the extremity of a given line (pl. 8, fig. 29, and pl. 2, fig. 
76). 9. To draw a given line between the sides of a given angle bac, so 
that it may form a given angle m with one of the two sides (pl. 3, fig. 30). 
10. To construct a triangle with a given line a, and two given angles m 
and n [the sum of which must be less than two right angles], (fig. 31). 
11. To construct an equilateral triangle upon a given base (pl. 2, fig. 78). 
12. To construct a square upon a given base ab ( pl. 2, figs. 79, 80). In order 
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