407. Of all polygons formed of sides all given but one, the polygon inscribed in a semicircle, having the undetermined side for its diameter, is the maximum.

Let AB, BC, CD, and DE be the sides of a polygon inscribed in a semicircle having AE for its diameter.

We are to prove the polygon ABCDE the maximum of polygons having the sides A B, BC, C D, and D E.

From any vertex, as C, draw CA and C E.

Then the

ACE is a rt. Z,

§ 204

(being inscribed in a semicircle).

Now the polygon is divided into three parts, ABC, CDE,

and A CE.

The parts ABC and CDE will remain the same, if the ZACE be increased or diminished;

but the part A CE will be diminished,

§ 406

(of all having two sides respectively equal, that in which these sides include a rt. is the maximum).

.. A B C D E is the maximum polygon.

Q. E. D.

PROPOSITION XXV. THEOREM.

408. The maximum of all polygons formed of given sides can be inscribed in a circle.

Let ABC DE be a polygon inscribed in a circle, and A'B'C' D'E' be a polygon, equilateral with respect to ABCDE, but which cannot be inscribed in a circle.

We are to prove

the polygon A B C DE> the polygon A'B'C' D' E'.

Draw the diameter A H.

Join CH and D H.

Upon C' D' (= C D) construct the ▲ C'H' D'ACHD,

and draw A' H'.

Now the polygon A B C H> the polygon A'B'C' H', § 407 (of all polygons formed of sides all given but one, the polygon inscribed in a semicircle having the undetermined side for its diameter, is the maximum). And the polygon A ED H> the polygon A' E' D' H'.

Add these two inequalities, then

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the polygon ABCH DE> the polygon A'B'C'H' D'E'. Take away from the two figures the equal ACHD and C'H'D'.

Then the polygon ABCDE> the polygon A'B'C' D' E.

Q. E. D.

409. Of all triangles having the same base and equal perimeters, the isosceles triangle is the maximum.

Let the AACB and ADB have equal perimeters, and let the ▲ ACB be isosceles.

(having the same base are to each other as their altitudes).

The

Produce AC to H, making C H = A C.

Draw HB.

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ABH is a rt. ≤, for it will be inscribed in the semicircle drawn from C as a centre, with the radius C B.

From Clet fall the LCK;

and from D as a centre, with a radius equal to D B,

Since

and

describe an arc cutting H B produced, at P.

(a drawn from the vertex of an isosceles ▲ bisects the base),

§ 56

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Q. E. D.

PROPOSITION XXVII. THEOREM.

410. The maximum of isoperimetrical polygons of the same number of sides is equilateral.

B K

D

Let ABC D, etc., be the maximum of isoperimetrical polygons of any given number of sides.

We are to prove AB, BC, CD, etc., equal.

Draw A C.

The AABC must be the maximum of all the A which are formed upon A C with a perimeter equal to that of ▲ A B C.

Otherwise, a greater ▲ A KC could be substituted for ▲ ABC, without changing the perimeter of the polygon.

But this is inconsistent with the hypothesis that the polygon ABCD, etc., is the maximum polygon.

.. the ABC, is isosceles,

$409 (of all having the same base and equal perimeters, the isosceles ▲ is the

maximum).

In like manner it may be proved that BC :

=

CD,

etc.

Q. E. D.

411. COROLLARY. The maximum of isoperimetrical polygons of the same number of sides is a regular polygon.

For, it is equilateral,

§ 410

(the maximum of isoperimetrical polygons of the same number of sides is

equilateral).

Also it can be inscribed in a O,

§ 408

(the maximum of all polygons formed of given sides can be inscribed in a O).

Hence it is regular,

(an equilateral polygon inscribed in a ○ is regular).

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