PLANIMETRY. ? 
to obtain a regular octagon from a square, we must proceed as follows :— 
Draw (pl. 3, fig. 32) the two diagonals of the quadrilateral intersecting 
ate: bisect the angle ced; make the bisecting line ef= ce or de, and draw 
cf and df; finally, describe upon the remaining three sides of the quadri- 
lateral, equilateral triangles, which are equal to the triangle cdf. 
4. OF THE SIMILARITY OF FIGURES. 
Two figures are called similar when they agree in their form, or more 
definitely, when the angles of the one are equal to the angles of the other 
each to each in the same order, and the sides of the one are in the same 
proportion as those of the other. We arrive at the latter definition as 
respects triangles, by examining two lines not parallel to each other, which 
are intersected by several parallel lines. If we divide one of the two lines 
aq, ar ( pl. 3, fig. 33), into the equal parts, ab = be = cd = de, commencing at 
the point of intersection, and then draw from the points of division, 8, ¢, d, 
e, parallel lines, then the divisions of the other lines thus formed will also be 
equal to each other. If, again, we take in one of the two lines aq, ar (fig. 
34), two or more unequal parts, beginning at the point of intersection a, and 
then draw parallel lines from the points of division 6, c, then will the 
resulting sections, ad, de, of the other line, be in the same proportion to one 
another as the sections, ab, bc. In this case the triangles abd, ace, are 
similar ; we readily perceive that their angles are equal, and that two sides 
of the one triangle are always in the same proportion as the corresponding 
and similarly placed sides of the other. To be certain that two given: 
triangles, abc, def (fig. 35), are similar, it is only necessary to know,—1, that 
two angles, or 2, that one angle and the ratios of the including sides, or 8, 
that the ratios of two of the sides and the angles opposite to the greater 
of them, or 4, that two ratios of sides, are equal. 
The following propositions, among many others, may be proved by 
means of the similarity of triangles. If in a triangle abc (pl. 3, fig. 36), lines 
be drawn from two angles, a, b, to the middle of the opposite sides, each of 
these lines cuts the other into two parts, of which the one lying towards 
the bisected side is half of the other. From this it readily follows that all 
the three lines drawn from the angles of a triangle to the middle of the 
opposite sides, pass through one and the same point. If we bisect 
the angle a of the triangle abc (pl. 3, fig. 37), it may be readily shown 
that the segments into which the opposite side bc is divided by the bisect- 
ing line, are in the same proportion as the two other undivided sides, thus, 
bd:ed::ab:ac. Wence we deduce the proposition, that lines bisecting the 
three angles of a triangle cut each other in one and the same point, which 
is of importance, as being the centre of the circle inscribed in the triangle. 
To cut off from a given triangle a smaller one similar to it, we may either, 
1, draw a line parallel to one side of the triangle, or 2, from one angle of 
the triangle, not the least, cut off by a line another angle equal to a 
smaller one of the same triangle. E. g., if in fig. 38 the angle n= m, then 
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