THEOREM C.

Of two regular polygons having the same perimeter, that which has the greater number of sides is the greater.

Place a side of one on a side of the other, so that their middle points coincide. Let B A be half the side of one, and 2m times B A the perimeter; and let BC be half the side of the other, and 27 times BC the perimeter; and let B A > BC. Also let O be the centre of that which has the greater side, and E the centre of that which has the less.

Draw O D parallel to E C, and with O as centre describe arc FDG.

Since the areas are respectively

half perimeter x BO and half perimeter x B E, if BE > BO the area of the first will be greater than that of the second. We shall show that BE > BO by proving that BC BD.

Because 2m x 4 BOA = 4 right angles,

and 2n xB EC 4 right angles; ..m: n = 4 BEC: 4BOA = 4BOD: BOA, sector FOD: sector F O G.

But since by hypothesis 2m B A = 2n · BC:

..m:n BC: BA;

.. sector F O G sector F O D=BA: BC;

.. sector DOG: sector FOD-CA:B C. But sector DO G: sector FOD < A DOA: A BOD, that is DA: BD;

therefore CA: CB< DA: DB,

BA:CB BA: DB;

.. BC > D B.

Hence also BE > BO,

and the area of the polygon on BC greater than that

on B A.

THEOREMS FOR EXERCISE.

123. Arcs of circles which subtend equal angles at the centre are to one another as the radii of the circles.

124. Of all rectangles having the same area the square has the least perimeter.

125. A B C is a triangle, and A D the line bisecting the angle B AC; through A the line A E is drawn at right angles to A D. Prove that if P be any point on the line A E, the perimeter of the triangle B PC will be greater than that of AB C.

126. Of all parallelograms having sides of given length the rectangle is the greatest.

127. Of all triangles having the same base and area the isosceles has the least perimeter.

128. Of all triangles on the same base, and having the same vertical angle, the isosceles has the greatest area.

MISCELLANEOUS THEOREMS AND PROBLEMS.

1. Two straight lines, CO, CM, are drawn from the vertex, C, of an angle, A CB, the first, CO, bisecting the angle and the second, CM, dividing it into two unequal parts. Prove that half the difference between AC M and BCM is equal to Z, MCO.

2. If the lines CM in the preceding exercise be drawn without the angle, then half the sum of the angles A CM, BC M will be equal to M CO.

3. The angle between the bisector of the angle A of ▲ A B C, and the perpendicular from A on B C is equal to half the difference of B and C.

4. If through any point on the base of an isosceles triangle parallels be drawn to the other two sides, prove that the parallelogram so found will have the same perimeter for all positions of the point.

5. Two quadrilaterals inscribed in a circle are equal in all respects when the four sides of the one are respectively equal to the four sides of the other taken in the same order.

6. Quadrilaterals which have their diagonals equal and inclined at the same angle are equal.

MISCELLANEOUS THEOREMS AND PROBLEMS.

241

7. The two straight lines which join the middle points of opposite sides of a quadrilateral and the straight line which joins the middle points of the diagonals of the quadrilateral pass through the same point, and each is bisected at the common point.

8. Inscribe in a given circle a given number of equal circles touching one another and the given circle.

9. If a chord of a circle be produced till the part produced be equal to the radius, and if from its extremity a line be drawn through the centre and meeting the convex and concave circumferences, the convex is one-third of the concave circumference.

10. From the preceding theorem devise a mechanical contrivance for trisecting a given angle.

II. Construct a quadrilateral, having given the four sides and the straight line which joins the middle points of the opposite sides.

12. The bisectors of the angles of a quadrilateral form a quadrilateral, of which the opposite angles are supplementary. When the first quadrilateral is a parallelogram the second is a rectangle; when the first is a rectangle the second is a square.

13. A ball is struck from a point on a rectangular billiard-board, and after rebounding from each side reaches the point of departure. The path of a ball before and after impact makes equal angles with the side: draw the path described.

14. The perpendicular bisectors of the three sides of a triangle meet in a point which is equidistant from the three vertices.

15. If E be the point of intersection of the perpendicular bisectors of the sides of a triangle, F the point of intersection of the three middle lines, and G the intersection of the three altitudes; E, F, and G lie in the same straight line.

16. A B C is a triangle inscribed in a circle. Show that the straight line joining the points in which the bisectors of the angles A and C meet the circumference of the circle cut off equal lengths from B A and B C.

17. A B C is a triangle having a right angle at C. Find a point, D, in BC so that the difference of the squares on A B and A D may be equal to three times the difference of the squares on A D and A C.

18. A D B is a triangle of which the angle is greater than the angle A, and AC B an isosceles triangle on the same side of A B as ▲ AD B, such that

If E be the point in which C B cuts A D prove that EA is greater than E B and E C greater than ED.

19. Construct a triangle, having given the lines drawn from the middle points of the sides to the opposite angles.

20. The three middle points of the sides of a triangle and the three

R

feet of the perpendiculars drawn from the vertices to the sides respectively opposite to them all lie on the circumference of a circle which also passes through the middle points of those portions of the three perpendiculars which lie between their common intersection and the vertices of the triangle.

21. The locus of the centres of circles inscribed in right-angled triangles on the same hypothenuse is the quadrant described on the hypothenuse.

22. A straight line is divided into two unequal parts: find the locus of points at which the two parts subtend equal angles.

23. Find the locus of points from which the tangents to either of two given circles will contain the same angle.

24. Prove that if common tangents be drawn to each pair of three given circles having unequal radii the three points of intersection of exterior common tangents lie in the same straight line.

25. Prove that each of the points of intersection of interior common tangents lies in the same straight line with two points of intersection of exterior common tangents.

26. Given the radii of the inscribed and circumscribed circles of an isosceles triangle to construct it.

APPENDIX.

EUCLID has so long been used as a Text-Book in our schools and colleges, that to many the following tabular view of the correspondence of his propositions with those of this work may be acceptable. The numbers in the left-hand column of the table indicate the propositions in Chambers's edition of Euclid, and those in the second column, their counterparts, as given in the present treatise. When a proposition in Euclid has no corresponding one in this work, the page on which it might be logically inserted as an additional exercise is indicated. It must be borne in mind, however, that, as the order of arrangement and mutual dependence of the propositions are different in the two cases, the demonstrations of corresponding propositions are seldom identical. One main source of this want of identity originates in the fact that Euclid rigidly adheres to the rule not to employ in the construction that may be needed for the demonstration of a theorem, any problem, however simple, until he has first shown how the construction can be made. Thus the acknowledged difficulty in his demonstration of the Fifth Proposition of the First Book might have been removed, had his scheme permitted him to assume that there was some one line that bisected an angle, before showing how such a line could be drawn. In this treatise, on the contrary, it is assumed that such a bisecting line exists, and may be represented ;* and the reasoning is not thereby vitiated. In like manner, it is taken for granted that a line has a middle point, before showing how to find that point: and so for other simple problems. In this way, theorems are rendered independent of problems, and their demonstrations are not only thereby simplified, but a more natural sequence of propositions can be observed.