MISCELLANEOUS PROPOSITIONS.

1. DESCRIBE a triangle in a circle whose angles are as 1, 2, and 3. SUGGESTION.-Divide the circumference into six equal parts, and draw chords.

2. Describe a triangle in a circle whose angles shall be as 3, 5, and 8.

3. Describe a triangle in a circle whose angles shall be as 2, 5, and 8.

4. Divide a circle into two segments such that the angle in one segment shall be twice the angle in the other.

5. Divide a circle into two segments such that the angle in one segment shall be five times the angle in the other.

6. If two equilateral triangles are described about a circle, their intersection forms an equilateral hexagon.

7. Construct a circle which shall touch a given straight line and bisect the circumference of a given circle.

8. Construct a circle which shall touch one side of a triangle and the other two sides produced.

NOTE. Such a circle is called an escribed circle.

9. Upon a given straight line as a base, construct an isosceles triangle having the third angle double each of the angles at the base. 10. Upon a given straight line as a base, construct an isosceles triangle having the third angle four times each angle at the base. 11. Upon a given straight line as a base, construct an isosceles triangle having the third angle six times each angle at the base.

12. If two straight lines whose extremities are in a circumference cut each other, the triangles formed by joining their extremities are mutually equiangular.

13. If two equal chords intersect in a circle, the segments of one are respectively equal to the segments of the other (IV. 31).

14. Draw two concentric circles such that the chords of the outer circle which are tangent to the inner circle may be equal to its diameter.

NOTE.-Several of the above propositions are selected from the examination papers of Cambridge University.

SUPPLEMENT.

1. THE SUPPLEMENT presents additional matter for those who wish to pursue the subject further.

2. It is divided into three parts, as follows:

I. MAXIMA AND MINIMA. II. SUPPLEMENTARY PROPOSITIONS. III. MISCELLANEOUS PROPOSITIONS.

SECTION I,

MAXIMA AND MINIMA.

1. Or all quantities of the same kind, that which is the greatest is the maximum; and that which is the least is the minimum.

Thus, the diameter of a circle is a maximum among all straight lines joining two points of the circumference. The perpendicular is a minimum among all the lines drawn from a point to a given straight line

2. Figures which have equal perimeters are called isoperimetric figures.

PROPOSITION I.-THEOREM.

Of all triangles having two sides respectively equal, that in which these sides include a right angle is the maximum.

Let ABC and ABD be two ▲'s having the sides AB and BC respectively equal to AB and BD, the angle ABC being a right angle; then the ABC is greater than the ▲ ABD. For, the A's are to each other as their altitudes (IV. 6, C.); but the altitude DE is less than BC (I. 6); hence

B

E

▲ ABC > ▲ ABD.

PROPOSITION II.-THEOREM.

Of all triangles having the same base and equal areas, that which is isosceles has the minimum perimeter.

Let ABC be an isosceles A, and DBC any other▲ having the same base and an equal area; then AB+AC is less than DB+DC.

For, the altitudes of the two triangles are equal (IV. 6); hence the vertices A and D lie in the same line DF parallel to BC.

=

B

= ABC ~

x

Draw CFE to DF, meeting BA produced in E; join DE. Since FAC= LACB= LEAF, the 's ACF and AEF are equal; hence AF is CE at its middle point F; and AE= AC, and DE=DC; but BE <BD+DE; therefore AB+ AC <DB + DC.

to

COR.-Of all triangles having the same area, the equilateral triangle has the minimum perimeter.

PROPOSITION III.-THEOREM.

Of all isoperimetric triangles having the same base, the isosceles triangle is the maximum.

Let ABC be an isosceles A, and let the A DBC on the same base BC, have an equal perimeter; then the area of ▲ ABC is greater than the area of ▲ DBC.

E..

B

Draw EF parallel to BC. The point D will fall between the parallels EF and BC; for if it fell on EF we should have (as in the preceding demonstration) BD + DC > ABAC; and if it fell above EF, the sum BD+ DC would be still greater; hence it falls below EF.

Therefore the altitude of ▲ ABC is greater than the altitude of ▲ DBC, and consequently area ▲ ABC area ▲ DBC.

COR.-Of all isoperimetric triangles, that which is equilateral is the maximum.

PROPOSITION IV.-THEOREM.

Of all isoperimetric figures, the circle is the max

imum.

E

D

1st. The maximum figure must be convex. If EACB is a figure not convex, we could substitute for the line ACB the convex line ADB of equal length, which would increase the area of the figure without increasing the perimeter. Hence a maximum figure must be convex. 2d. Suppose ACBDA to be a maximum figure formed with a given perimeter. Draw AB bisecting the perimeter; then AB will also bisect the area.

If not, suppose ACB greater than ADB; then a line equal to ACB could be substituted for ADB, which would, with an equal perimeter, cut off a greater

area.

But ACBD was, by hypothesis, the maximum; hence the area is bisected by AB.

E

Again, take any point C in the perimeter

and draw AC and CB; then the angle ACB is a right angle. For, if not, draw CE at right angles to CB; make it equal to AC, and on it describe a segment equal to the segment on AC.

Then ▲ ECB > ▲ ACB (Th. I.); hence the whole figure EBGCF is greater than the whole figure ABGCH. Now, if figures respectively similar to those on the upper side of EB and AB, be described on the opposite sides of these lines, the entire figure on EB will be greater than the entire figure on AB; and the figures have equal perimeters.

Hence the hypothesis that the angle ACB is not a right angle leads to the conclusion that the figure ACBD is not a maximum, which is contrary to the hypothesis; therefore the angle ACB must be a right angle.

But B is any point of the perimeter ACBD; hence the curve ACB is a semi-circumference, and the figure ACBD is a circle, and AB is its diameter.

PROPOSITION V.-THEOREM.

Of all plane figures containing the same area, the circle has the minimum perimeter.

Let A be a circle, and B be any other figure having the same area as A; then the perimeter of A is less than that of B.

For, let C be a circle having the same perimeter as the figure B. Then (Th. 4) C> B, but B=A. Hence, C > A.

Therefore, circ. A circ. C.

But circ. C=per. B. Hyp.
Hence,

circ. A per. B.

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PROPOSITION VI.-THEOREM.

Of all polygons having equal sides, that is the maximum which can be inscribed in a circle.

For upon the sides of the polygon P' construct circular segments respectively equal to the segments on the corresponding sides of P; the whole figure C' thus formed has the same perimeter as the circle C; hence the area of C, being a circle, is greater than the area of C' (Th. IV.); and subtracting the equal circular segments from both, we have the area of P greater than the area of P'.