A'B'C D', the area of the circle ABCD > A'B'C'D' (Ex. 2), but the segments AB=A'B', BC=B'C', etc.: take away these equal parts and there remains the quadrilateral ABCD greater than A'B'C'D'.]*

4. If three sides a, b, c, of a quadrilateral are given in magnitude, the area is a maximum when the fourth side d is the diameter of the circle through the vertices; and generally,

When all the sides but one of a polygon of any order are given in magnitude, the area is a maximum when the circle on the closing side as diameter passes through the remaining vertices.† [Proof as above.]

5. Having given of a quadrilateral the diagonals & and 8' and a pair of opposite sides BC and AD, its area is a maximum when BC is parallel to AD.

[Take any position of the quadrilateral and through C draw CE parallel and equal to 8. Join BE and AE.

The triangles BDE and BCD are equal (Euc. I. 37); to each add ABD, therefore ABCD=ABED.

*The student should learn the proof of the Trigonometrical expression for the area of any quadrilateral in terms of the four sides and the sum of either pair of opposite angles.

(Area)2 = (s − a)(s – b)(s — c)(s – d) – abcd cos2(A + C).

(Casey's Plane Trig., art. 152, cors. 3, 4.)

+ To construct the quadrilateral. Let 0 be the angle between a and b, and AC = x.

Then

but

d2 = c2+x2= a2+b2+c2 - 2ab cos 0 ;............

cos 0 = -c/d;

.(1)

substituting in (1) and simplifying we have the following expression for d:

d3 - d(a2+b2+c2) - 2abc = 0,

an equation which has only one positive root. (Burnside and Panton's Theory of Equations, Art. 13.)

In the particular case when a = b = c, the equation for d reduces to

thus showing that the quadrilateral is half the regular inscribed hexagon.

Now, ABDE is a maximum when AD and DE are in the same straight line; hence ABCD is a maximum when BC is parallel to AD.]

6. The diagonals of a quadrilateral are 9 and 10 feet and two opposite sides 5 and 3 feet; find when its area is a maximum.

10. Theorem. Having given the base AB of a triangle and the locus of the vertex a line L meeting the base produced, the sum of the sides AC+ BC is a minimum when L is the external bisector of the vertical angle.

Let fall a perpendicular BL and make B'L-BL. Join AB and let C be its intersection with L. Take any other point P on the line and join AP and B'P.

The triangles BCL and B'CL are equal in every respect (Euc. I. 4); hence BC= B'C. Similarly BP = B'P. Hence since (Euc. I. 20) AP+B'P> AB' it follows that AP+BP > AC + BC.

COR. 1. If the line L cuts the base internally the difference of the sides (AC-BC) is a maximum when it bisects internally the angle C.

EXAMPLES.

1. The triangle of minimum perimeter inscribed in a given one is formed by joining the feet X, Y, Z of the perpendiculars* let fall from the vertices on the opposite sides.

[For the joining lines are equally inclined to sides on which they intersect (Euc. III. 21).]

2. The polygon of least perimeter that can be inscribed in a given one is that whose angles are bisected externally by its sides. (By Ex. 1.)

3. The base and area of a triangle being given, the perimeter is least when the triangle is isosceles.

[For the line L is parallel to the base.]

4. If from 0, the point of intersection of the diagonals of a cyclic quadrilateral, perpendiculars are drawn to the sides and their feet P, Q, R, S joined, the quadrilateral PQRS is of minimum perimeter.

4a. If points P', Q, R', S' be taken on the sides of the given quadrilateral, such that P'Q', Q'R', R'S' are parallel to PQ, QR, RS, then P'S' is parallel to PS and the perimeters of the quadrilaterals are equal. [Euc. VI. 2 and I. 5.]

5. The value of the minimum perimeter of the indeterminate inscribed quadrilateral in Ex. 4 is 288′/D, where D is the diameter of the circum-circle.

6. Given a triangle ABC, find a point O such that

OA+OB+OC is a minimum.

[Where BOC=COA=AOB=120°.]

* These are generally known as the Perpendiculars of the Triangle, and XYZ as the Pedal Triangle of ABC.

11. Problem.-Given an angle C of a triangle and a point P on the base, construct the triangle of minimum

Through P draw APB such that AP-BP. triangle ABC is less than any other A'B'C.

The

For draw AX parallel to BB'. Then the triangles APX and BPB' are equal in all respects (Euc. I. 4); hence AA'P > BB'P. To each add APB'C, therefore A'B'C>ABC; hence the triangle of least area is that whose base is bisected at this point.

12. Theorem. Given an angle and any curve concave to its vertex C. The tangent AB which forms with the sides of the angle a triangle ABC of minimum area is bisected at its point of contact (P).

For this tangent cuts off a less area than any other line A'B' through P, because it is bisected at P. Now draw any other tangent XY, and let PA'B' be parallel to it. Since the curve is concave to C, A'B'C< XYC; a fortiori ABC <XYC.

COR. 1. In the particular case when the curve is a circle whose centre is at C the triangle is isosceles.

property may be stated otherwise. When the vertical angle and altitude of a triangle are given, the base and area are both minima when the triangle is isosceles.

On account of its importance an independent proof of this property of the isosceles triangle is given.

Let ABC be an isosceles triangle and A'B'C any other, having the same vertical angle and altitude CM.

Now BC> B'C (Euc. III. 8), but BC=AC<A'C, hence A'C> B'C. Let CD=B'C, join AD. The triangles ACD and BB'C are equal in every respect (Euc. I. 4), hence AA'C> BB'C; therefore A'B'C > ABC; therefore, etc.

COR. 2. When the curve is a circle touching the sides of the angle the tangent AB and area ABC are each minima when the triangle is isosceles.