Proposition 6.

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

1. The maximum polygon must be equilateral; for, if not, an isosceles triangle, ABC, could be substituted for ADC of equal perimeter, and which would be greater (Prop. 2), and thus the area of the polygon would be increased without changing the perimeter.

B

D

C

2. The maximum polygon must be such as to be inscribed in a circle. For, let P, P' be equilateral polygons of the

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.

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

A

C B

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

SECTION IV.

TRANSVERSALS.

DEFINITIONS.

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.

3. The diagonals of a complete quadrilateral are the three lines joining the opposite angles-viz. DB, AC, EF.

A

F

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.

с

B

E

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

Α

C B

AC : CB :: ᎪᎠ : ᎠᏴ,

or, which is the same, when

AC AD: BC: BD.

Ꭰ

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.

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

also in the similar triangles CDb, Abc, CD Cb Ac: Ab;

multiplying together the corresponding B

terms,

hence,

A

aC Cb: aB. Ac: Bc. Ab;

a C.bA.cBaB.bC.cA.

C

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.

Let the three points a, b, c be taken in the sides of ABC, so that

A

a

; but this can only be true

when coincides with c; hence abc is a straight line.

Proposition 10.

Theorem.-The three lines drawn from any point in the plane of a triangle to the three angles, and cutting the opposite sides (produced if necessary), will cut them so that the product of three non-adjacent segments is equal to the product of three others.

Let the three lines, A Oa, BOb, COc, be drawn through the angles of ABC and any point, O, in its plane,

then aB.bC.cA a C. bA.cB.

=

Because the triangle A Ca is cut by the transversal Bb,

(Prop. 8), Ba. OA.bC-BC. Oa.bA.

Also because the triangle ABa is cut by the transversal Cc,

CB. Oa. cA Ca. OA.cB.

=

Multiplying these together, we have

aB.bC.cA aC.bA.cB.

=

B

a

Proposition 11.

Theorem.-If three points be taken on the sides of a triangle, so that the product of three non-adjacent segments is equal to the product of the other three, the three straight lines joining these points with the opposite angles will pass through the same point.

The proof is similar to (Prop. 9).

From the last theorem the following propositions may be proved :

1. The three lines drawn from the angles of a triangle to the middle points of the opposite sides intersect in the same point. 2. The three lines drawn bisecting the angles of a triangle intersect in the same point.

3. The three perpendiculars drawn from the angles of a triangle to the opposite sides intersect in the same point.

Proposition 12.

Theorem.—In a complete quadrilateral each diagonal is divided harmonically by the other two.

Let the diagonal EF be cut by the diagonals AC, BD in P and Q; then will

PE PF: QE : QF.

Because the transversal QDB cuts the triangle AEF, we have (Prop. 8)