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THE elements of

and inscribed in the same circle; thatis the greatest whose sides are all equal.—(PI. 4, fig, 6.)

For, if possible, let some polygon ABC FE,whose sides C F, FE, are unequal, be thegreatest.

Let C D E be an isosceles triangle describedin the fame segment with C F E, and thro'C E and F describe the equal parallelsCE, FI. (P.L B. II.)

Because the point D lies above F, it willalso lie above any other point in the parallelIF. Let O, therefore, be the point whereIF cuts C D and join O E.

The triangle C D E being greater thanC O E, is also greater than its equal CFE.(C. I. T. V. B. II.) Whence, the whole polygonA B C D E will be greater than A B C F E,which is absurd. Therefore, the polygon isthe greatest when the fides are all equal.

THEOREM XV.

Of all figures on the sphere containedunder an equal perimeter of great