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But

BD <A'B + A'D,

(156. Any two sides of a triangle are together greater than the third.)

:. AB + AC < A'B + AC.

440. COROLLARY. Of all equivalent triangles, that which is equilateral has the least perimeter.

For the triangle having the least perimeter enclosing a given surface must be isosceles whichever side is taken as the base.

THEOREM II.

441. Of all isoperimetric triangles having the same base, that which is isosceles has the greatest surface.

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HYPOTHESIS. Let ABC be an isosceles triangle; and let A'BC standing on the same base BC, have an equal perimeter; that is, A'B + A'C = AB + AC.

CONCLUSION. Δ ABC > Δ ΑΒC.

PROOF. The vertex A' must fall between BC and the parallel AN; since, if it fell upon AN, by the preceding proof, A'B + A'C > AB + AC; and, if it fell beyond AN, the sum A'B + A'C would be still greater.

Therefore the altitude of ▲ ABC is greater than the altitude of ▲ A'BC, and hence also its surface.

442. Corollary. Of all isoperimetric triangles, that which is equilateral is the greatest.

For the greatest triangle having a given perimeter must be isosceles whichever side is taken as the base.

THEOREM III.

443. Of all triangles formed with the same two given sides, that in which these sides are perpendicular to each other has the greatest surface.

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HYPOTHESIS. Let ABC, A'BC, be two triangles having the sides AB, BC, respectively equal to A'B, BC; and let ABC be right. CONCLUSION. Δ ABC > Δ ΑΒC.

PROOF. Taking BC as the common base, the altitude AB of ▲ ABC is greater than the altitude A'D of ▲ A'BC.

THEOREM IV.

444. Of all isoperimetric plane figures, the circle contains the greatest surface.

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B

PROOF. With a given perimeter, there may be an indefinite number of figures differing in form and size. The surface may be as small as we please, but cannot be increased indefinitely.

Therefore, among all the figures of the same perimeter, there must be one greatest figure, or several equivalent greatest figures of different forms.

Every closed figure, if the greatest of a given perimeter, must be convex; that is, such that any sect joining two points of the perimeter lies wholly within the figure. For let ACBNA be a non-convex figure, the sect AB, joining two of the points in its perimeter, lying without the figure; then if the re-entrant portion ACB be revolved about the line AB into the position AC'B, the figure AC'BNA has the same perimeter as the first figure, but a greater surface.

Now let AFBCA be a figure of greatest surface formed with a given perimeter; then, taking any point A in its perimeter, and drawing

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AB to bisect the perimeter, it also bisects the surface. For if the surface of one of the parts, as AFB, were greater than that of the other part, ACB, then if the part AFB were revolved upon the line AB into the position AF'B, the surface of the figure AF'BFA would be greater than that of the figure AFBCA, and yet would have the same perim

eter.

Now the angles AFB and AF'B must be right angles, else the triangles AFB and AF'B could be increased, by 443, without varying the chords AF,FB, AF',F'B, and then (the segments AGF, FEB, AG'F', F'E'B, still standing on these chords) the whole figure would have increased without changing its perimeter.

But F is any point in the curve AFB; therefore this curve is a semicircle.

(384. The arc of a segment is a semicircle if the angle in it is right.)

Therefore the whole figure is a circle.

THEOREM V.

445. Of all equivalent plane figures, the circle has the least perimeter.

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HYPOTHESIS. Let C be a circle, and A any other figure having the same surface as C.

CONCLUSION. The perimeter of C is less than that of A.

PROOF. Suppose B a circle with the same perimeter as the figure A; then, by 444, A <B

..

C< B.

But, of two circles, that which has the less surface has the less perimeter ;

.. perimeter of C < perimeter of B, or of A.

THEOREM VI.

446. Of all the polygons constructed with the same given sides, that is the greatest which can be inscribed in a circle.

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HYPOTHESIS. Let P be a polygon constructed with the sides

a, b, c, d, e, and inscribed in a circle S, and let P' be any other

polygon constructed with the same sides, and not inscriptible in a circle.

CONCLUSION. P> P'.

PROOF. Upon the sides a, b, c, etc., of the polygon P' construct circular segments equal to those standing on the corresponding sides of P. The whole figure S' thus formed has the same perimeter as the circle S, therefore, by 444, surface of S> S'; subtracting the circular segments from both, we have

P> P'.

THEOREM VII.

447. Of all isoperimetric polygons having the same number of sides, the regular polygon is the greatest.

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PROOF. I. The greatest polygon P, of all the isoperimetric polygons of the same number of sides, must have its sides equal; for if two of its sides, as AB', B'C, were unequal, we could, by 439, increase its surface by substituting for the triangle AB'C the isoperimetric isosceles triangle ABC.

II. The greatest polygon constructed with the same number of equal sides must, by 446, be inscriptible in a circle. Therefore it is a regular polygon.

EXERCISES. 87. Of all triangles that can be inscribed in a given triangle, that whose vertices are the feet of the altitudes of the original triangle has the least perimeter.

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