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Proposition 3. Theorem.-Of all triangles having two sides equal, each to each, that which has these sides at right angles is the maximum.

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Let AB, BC, at right angles to each other, be equal, each to each, to DB, BC, not at right angles to each other; ABC is greater than DBC.

Draw the altitude DE; AB is greater than DE; hence ABC is greater than DBC.

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Proposition 4. Theorem.-Of all isoperimetric figures, the circle is the maximum.

1. The given maximum figure must be convex; that is, a line joining any two points of the perimeter must fall inside the perimeter. For, if EACB be a figure not convex, for the part ACB could be substituted the convex line ADB of equal length, which would increase the area of the figure without increasing the perimeter. Hence, any figure not convex cannot be a maximum.

2. Let ABCD be the maximum figure. Draw A C bisecting the perimeter; it will also bisect the area. For, if not, let ABC be greater than ADC; then a line equal to ABC could be substituted for ADC, which would, with an equal perimeter, cut off a greater area. But ABCD was, by hypothesis, the

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maximum. Hence, the area is bisected by AC

Also, take any point B in the perimeter; the angle ABC is a right angle. For, if not, draw BE at right angles to BC, making it equal to BA, and on it describe a segment equal to the segment on BA. Then (Prop. 3) the triangle EBC is greater than the triangle ABC; hence, the whole figure EFBGC is greater than the whole figure AHBGC; and if symmetrical figures be described on the opposite sides of EC and AC, the figure on EC would be greater than the figure on AC. But they have equal perimeters. Hence, ABCD is not a maximum, which is contrary to the hypothesis. Therefore, ABC is a right angle.

Now, B is any point of the perimeter; hence, ABCD is a circle, and AC its diameter.

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Proposition 5. Theorem.-Of all equal figures, the circle has the minimum perimeter.

Let A be a circle, and B any other equal figure; then the perimeter of A will be less than the perimeter of B.

Let C be a circle having the same perimeter as B; then (Prop. 4) C>B; ..C>A; hence, the circumference of C is greater than the circumference of A ; therefore, also, the perimeter of B is greater than the circumference of A.

Proposition 6.

Theorem. - Of all isoperimetric polygons having the same number of sides, the regular polygon is the maximum.

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2. The maximum polygon must be such to be inscribed in a circle. For, let P, P be equilateral polygons of the same number of sides, of which P is inscribed in a circle. On the sides of P' describe segments of circles A'B'C', etc., equal to the segments ABC, etc. The whole figure A'B'C", etc., thus formed, has the same perimeter as the circle about P, and is therefore (Prop. 4) less than the circle. But the segments being equal, P must be greater than P'.

Hence, the maximum polygon, being equilateral and inscriptible, is regular.

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Proposition 7. Theorem.- Of all isoperimetric polygons, that which has the greatest number of sides is the greatest.

Let P, Q be regular polygons of four and five sides, and

Then

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having equal perimeters. Take any point C in AB. P may be considered as an

B irregular polygon of five sides, AC, CB being two. Hence (Prop. 6), P is less than Q.

In the same manner, Q may be proved less than an isoperimetric regular polygon of six sides, and so on.

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SECTION IV.
TRANSVERSALS.

DEFINITIONS.

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1. A transversal is a line cutting a system of lines.

2. A complete quadrilateral is the figure formed by four straight lines intersecting in six points, as ABCDEF.

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4. If a transversal cut the three sides of a triangle (produced if necessary), there are formed six segments, of which any two about an angle, as Ac, Ab, or Ba, Bc, are said to be adjacent, and any other two, as Ab, Ba, non-adjacent.

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5. A line is divided harmonically when it is divided internally and externally in the same ratio; thus AB is divided harmoni

A cally when

AC: CB :: AD: DB, or, which is the same, when

AC: AD :: BC: BD.

6. C and D are said to be harmonic conjugates, as also A and B.

Proposition 8. Theorem.-If a transversal cut the three sides of a triangle, the product of three non-adjacent segments is equal to the product of the other three.

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Let the transversal abc cut the three sides of the triangle ABC.

Draw CD parallel to AB.
Then in the similar triangles a CD, a Bc,

aC : CD :: aB : Bc;
also in the similar triangles CDb, Abc,

CD : Cb :: Ac : Ab; multiplying together the corresponding B terms,

aC : Cb :: aB. Ac : Bc. Ab; hence,

aC.bA.cB = ab.bC.CA.

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Proposition 9. Theorem.-If in the sides of a triangle (produced if necessary) three points be taken, so that the product of three nonadjacent segments be equal to the product of the other three, these points will be in one straight line.

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