The last tangent must form in like manner the angle a with the first tangent, otherwise the sum of the n exterior angles of the polygon would not equal four right angles.

$ 69 Since the exterior angles are all equal, their supplementary angles, the angles of the polygon, must be all equal.

Hence the polygon is equiangular and of n sides.

Q. E. D.

1019. The circumference of a convex closed curve is the limit which the perimeters of a series of inscribed and circumscribed polygons approach when the number of their sides is indefinitely increased; and the area of the curve is the limit of the areas of these polygons.

TO PROVE I. Its circumference is the common limit which the perimeters of a series of inscribed and circumscribed polygons approach when the number of their sides is indefinitely increased.

II. The area of the curve is the common limit which the areas of the inscribed and circumscribed polygons approach when the number of their sides is indefinitely increased.

Draw AB of any assigned length, no matter how small. We shall prove that the difference between the inscribed and circumscribed perimeters can be made less than AB.

Conceive the straight line AD, equal in length to the circumference of C, to be drawn perpendicular to AB. Join BD.

Now divide 4 right angles into such a number of equal parts that each part a shall be less than the angle BDA. Let n be the number of parts.

Circumscribe about C an equiangular polygon of ʼn sides whose exterior angle is a.

§ 1018 Join the points of tangency, forming an inscribed polygon of n sides.

From the vertices of the circumscribed polygon draw perpendiculars to the sides of inscribed polygon, thus forming 27 right triangles.

The angles at S, T, U, V, etc., between the tangents and chords, as x or y, are each less than a (§ 59), and therefore still less than angle ADB.

Of these angles x, y, etc., select the greatest, and place the right triangle containing it within the angle ADB in the position DEH.

B

In like manner place all the other right triangles along DA, as EFG, etc., irrespective of order. Let the last one extend to A'.

Thus the sum of the bases, or DA', equals the inscribed perimeter (less than DA), and the sum of the hypotenuses equals the circumscribed perimeter.

Produce DH to meet the perpendicular A'B' at K.

I. Any hypotenuse, as MP, can easily be proved less than M'P', the portion of DK included between perpendiculars at M and N. $99 Hence, adding all such inequalities, DK is greater than the sum of the hypotenuses DH, EG, etc.

That is, DK is greater than the circumscribed perimeter.

Now DA' is equal to the inscribed perimeter.

Hence the difference of the circumscribed and inscribed perimeters

is less than DK—DA'.

But DK-DA' is less than KA'.

§ 137

And KA' is less than A'B'. And A'B' is less than AB. Much more, therefore, is the difference of the perimeters less than AB.

We can thus make the difference of perimeters as small as we please. But the circumference is always intermediate between the perim

eters.

Hence either perimeter can be made to differ from the circumference by less than any assigned quantity.

Therefore the circumference is the common limit to which the perimeters approach.

Q. E. D.

II. Moreover, the difference between the areas of the inscribed and circumscribed polygons consists of the 2n right triangles, which is less than the triangle ADB.

But since the base AD of this triangle is constant we can make its area as small as we please by making its altitude AB as small as we please.

$187

Hence the difference between the polygons can be made as small as we please.

But the area of the curve is always intermediate between the polygons.

Hence either polygon can be made to differ from the area of the curve by less than any assigned quantity.

Therefore the area of the curve is the common limit to which the polygon-areas approach.

Q. E. D.

MAXIMA AND MINIMA OF PLANE FIGURES

1020. Def. Of the values which a variable quantity assumes, the largest value is called the maximum; the smallest, the minimum.

Thus, the diameter of a circle is the maximum among all straight lines joining two points of the circumference; and among all the lines drawn from a given point to a given straight line the perpendicular is the minimum.

1021. Of all triangles having the same base and equal areas, that which is isosceles has the minimum perimeter.

GIVEN the isosceles triangle ABC and any other triangle DBC having an equal area and the same base BC.

TO PROVE the perimeter of ABC is less than the perimeter of DBC.

Outline proof.-The vertices A and D are in the straight line XY parallel to BC. (Why?)

Draw CE perpendicular to BC, meeting BA produced at F. Join DF.

The angle CAE= angle FAE, and the triangle CAE triangle FAE. Hence AE is perpendicular to CF at its middle point.

1022. Remark.-The converse of the preceding proposition is also true, viz. Of all triangles having the same base and equal areas, that which has the minimum perimeter is isosceles. In fact, it is practically the same theorem as the proposition itself, for there is only one isosceles triangle fulfilling the given conditions, and only one triangle of minimum perimeter fulfilling the given conditions; just as to say that John Smith is the tallest man in the room is equivalent to saying that the tallest man in the room is John Smith, provided we know that there is only one John Smith in the room and only one tallest man. 1023. Cor. Of all triangles having the same area, that which is equilateral has the minimum perimeter.

1024. Def.-When two figures have equal perimeters they are called isoperimetric.

PROPOSITION IX. THEOREM

1025. Of all isoperimetric triangles having the same base, that which is isosceles has the maximum area.

GIVEN―the isosceles triangle ABC and any other triangle DBC having an equal perimeter and the same base BC.

TO PROVE

the area of ABC > area DBC.

Outline proof.-Draw AE perpendicular to BC and DF parallel to

Therefore the area of triangle ABC > area of triangle DBC. Q. E. Ü.