And the chords AB, BC, CD, etc., being sides of congruent equilateral triangles are all equal.

Therefore ABCDEF is a regular hexagon.

Cor. Since AOB is an equilateral triangle, AB=AO ; .. the side of a regular hexagon is equal to the radius of its circumcircle.

135°. Problem.-To determine which species of regular polygons, each taken alone, can fill the plane.

That a regular polygon of any species may be capable of filling the plane, the number of right angles in its internal angle must be a divisor of 4. But as no internal angle can be so great as two right angles, the only divisors, in 133°, Cor., are, I, and, which give the quotients 6, 4, and 3.

Therefore the plane can be filled by 6 equilateral triangles, or 4 squares, or 3 hexagons.

It is worthy of note that, of the three regular polygons which can fill the plane, the hexagon includes the greatest area for a given perimeter. As a consequence, the hexagon

is frequently found in Nature, as in the cells of bees, in certain tissues of plants, etc.

Ex. 1. Let D, E, F be points of contact of the inR1 circle, and P, P', P′′, R, R', R", etc., of the ex-circles. (131°)

Then AP AP',CP'=CP",

and BP BP", (114°,Cor. 1) ... AP'+AP

=AB+BC+AC =a+b+c,

and, denoting the perimeter of the triangle by 2s, we have

AP=AP'=s,

CP's-b=CP", BP s-c=BP".

Similarly, AR=s-b=AR", BR'=s-a=BR", etc.

Again,

Similarly,

CD=CE=6-AE=b-AF=b-(c-BF)

=b-c+BD=b-c+a− CD,

2CD=b+a-c=2(s — c),

CE CD=s-c=BP".

AE

AF s-a=BR", etc.

These relations are frequently useful.

If we put Aż to denote the distance of the vertex A from the adjacent points of contact of the in-circle, and Ab, Ac to denote its distances from the points of contact of the ex-circles upon the sides b and c respectively, we have

Ai=Bc=Cb=s-a,

Bi=Ca=Ac=s—b,

Ci=Ab= Ba=s-c.

EXERCISES.

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I. In testing the straightness of a rule" three rules are virtually tested. How?

2. To construct a rectangle, and also a square.

3. To place a given line-segment between two given lines so as to be parallel to a given line.

4. On a given line to find a point such that the lines joining it to two given points may make equal angles with the given line.

5. To find a point equidistant from three given points.

6. To find a line equidistant from three given points. How many lines?

7. A is a point on line L and B is not on L. To find a point P such that PA±PB may be equal to a given segment. 8. On a given line to find a point equidistant from two given points.

9. Through a given point to draw a line which shall form an isosceles triangle with two given lines. How many solutions?

10. Through two given points on two parallel lines to draw two lines so as to form a rhombus.

II. To construct a square having one of its vertices at a given point, and two other vertices lying on two given parallel lines.

12. Through a given point to draw a line so that the intercept between two given parallels may be of a given length. 13. To construct a triangle when the basal angles and the altitude are given.

14. To construct a right-angled triangle when the hypothenuse and the sum of the sides are given.

15. To divide a line-segment into any number of equal parts. 16. To construct a triangle when the middle points of its

sides are given.

17. To construct a parallelogram when the diagonals and one side are given.

18. Through a given point to draw a secant so that the chord intercepted by a given circle may have a given length. 19. Draw a line to touch a given circle and be parallel to a given line. To be perpendicular to a given line. 20. Describe a circle of given radius to touch two given lines. 21. Describe a circle of given radius to touch a given circle and a given line.

22. Describe a circle of given radius to pass through a given point and touch a given circle.

23. Describe a circle of given radius to touch two given circles. 24. To inscribe a regular octagon in a circle.

25. To inscribe a regular dodecagon in a circle.

26. A, B, C, D,

are consecutive vertices of a regular octagon, and A, B', C', D', ..., of a regular dodecagon in the same circle. Find the angles between AC and B'C'; between BE' and B'E. (Use 108°.)

27. Show that the plane can be filled by

(a) Equilateral triangles and regular dodecagons.
(b) Equilateral triangles and squares.

(c) Squares and regular octagons.

PART II.

PRELIMINARY.

136°. Def. I.- -The area of a plane closed figure is the portion of the plane contained within the figure, this portion being considered with respect to its extent only, and without respect to form.

A closed figure of any form may contain an area of any given extent, and closed figures of different forms may contain areas of the same extent, or equal areas.

Def. 2.-Closed figures are equal to one another when they include equal areas. This is the definition of the term equal" when comparing closed figures.

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Congruent figures are necessarily equal, but equal figures are not necessarily congruent. Thus, a ▲ and a □ may have equal areas and therefore be equal, although necessarily having different forms.

137°. Areas are compared by superposition. If one area can be superimposed upon another so as exactly to cover it, the areas are equal and the figures containing the areas are equal. If such superposition can be shown to be impossible the figures are not equal.

In comparing areas we may suppose one of them to be divided into any requisite number of parts, and these parts to be afterwards disposed in any convenient order, since the whole area is equal to the sum of all its parts.

B

A

Illustration.-ABCD is a square.

Then the ABC=^\ADC, and they are therefore equal.

D

Now, if AD and DE be equal and in line, the As ADC and EDC are congruent and equal.

Therefore the ABC may be taken E from its present position and be put into And the square ABCD is thus transformed into the ▲ACE without any change of area;

the position of CDE.

☐ABCD=^ACE.

It is evident that a plane closed figure may be considered from two points of view.

1. With respect to the character and disposition of the lines which form it. When thus considered, figures group themselves into triangles, squares, circles, etc., where the members of each group, if not of the same form, have at least some community of form and character.

2. With respect to the areas enclosed.

When compared from the first point of view, the capability of superposition is expressed by saying that the figures are congruent. When compared from the second point of view, it is expressed by saying that the figures are equal.

Therefore congruence is a kind of higher or double equality, that is, an equality in both form and extent of area. This is properly indicated by the triple lines (=) for congruence, and the double lines (=) for equality.

138°. Def.-The altitude of a figure is the line-segment which measures the distance of the farthest point of a figure from a side taken as base.

The terms base and altitude are thus correlative. A triangle may have three different bases and as many corresponding altitudes.

(87°)

In the rectangle (82°, Def. 2) two adjacent sides being perpendicular to one another, either one may be taken as the