Ex. 15. Transform a parallelogram into an equivalent parallelogram having a given base, but the angle adjacent to the base unchanged. Ex. 16. Transform a given triangle into an equivalent right triangle having a given leg.

[SUG. Use Ex. 14, then Ex. 5.]

Ex. 17. Transform a given triangle into an equivalent right triangle having a given hypotenuse.

[SUG. Find the altitude upon the hypotenuse of the new triangle by finding the fourth proportional to what three lines?]

Ex. 18. Transform a re-entrant pentagon into an

equivalent triangle.

Ex. 19. Transform a given triangle into an equivalent equilateral triangle.

[SUG. See Art. 416.]

Ex. 20. Bisect a triangle by a line perpendicular to one of its sides. [SUG. See Art. 416.]

Ex. 21. Construct a square equivalent to two-thirds of a given square.

EXERCISES. GROUP 44

PROBLEMS SOLVED BY ALGEBRAIC ANALYSIS

Ex. 1. Transform a given rectangle into an equivalent rectangle with a given base.

Ex. 2. Transform a given square into a right triangle having a given leg.

Ex. 3. Transform a given triangle into an equivalent isosceles right triangle.

Ex. 4. Draw a line cutting off from a given triangle an isosceles triangle equivalent to one-half the given triangle.

[SUG. Use Art. 397.]

Ex. 5. Through a given point in one side of a given triangle draw a line bisecting the area of the triangle.

Ex. 6. Transform a given square into a rectangle which shall have three times the perimeter of the given square,

EXERCISES. GROUP 45

PROBLEMS SOLVED BY VARIOUS METHODS

Ex. 1. Bisect a given parallelogram by a line parallel to the base.

Ex. 2. Transform a parallelogram into an equivalent parallelogram having the same base and a given side adjacent to the base.

Ex. 3. Construct a square which shall contain four-sevenths of the area of a given square.

Ex. 4. In two different ways construct a square having three times the area of a given square.

Ex. 5. Trisect a given triangle by lines parallel to the base.

Ex. 6. Find a point within a triangle such that lines drawn from it to the vertices trisect the area.

Ex. 7. Find a point within a triangle such that lines drawn from it to the three vertices divide the area into parts which shall have the ratio 2:3:4.

[SUG. If one of the small contains the area of original ▲, a line through its vertex cuts off the altitude, etc.]

Ex. 8. Divide a triangle into three equivalent parts by lines through a given vertex.

Ex. 9. Divide a triangle into three equivalent parts by lines drawn through a given point P in one of the sides.

[SUG. Use Art. 397.]

Ex. 10. Divide a given quadrilateral into three equivalent parts by lines drawn through a given vertex.

Ex. 11. Through a given point in the base of a trapezoid draw a line bisecting the area of the trapezoid.

Ex. 12. Bisect the area of a trapezoid by a line drawn parallel to the bases.

[SUG. Construct ▲ GEF similar to A ABG and equivalent tosum of what two A]

A

E

D

Book V

REGULAR POLYGONS. MEASUREMENT OF

THE CIRCLE

417. DEF. A regular polygon is a polygon that is both equilateral and equiangular.

PROPOSITION I. THEOREM

418. An equilateral polygon that is inscribed in a circle is also equiangular and regular.

B

K

Given ABC... K an inscribed polygon, with its sides. AB, BC, CD, etc., equal.

To prove the polygon ABC...K equiangular and regular.

Proof. Arc AB=arc BC=arc CD, etc.

.. arc ABC=arc BCD=arc CDE, etc.

:. Z ABC = 4 BCD = 2 CDE, etc.,

Art. 218.

Ax. 2.

Art. 260.

(all inscribed in the same segment, or in equal segments, are equal). .. the polygon ABC... K is equiangular.

..the polygon ABC... K is regular.

Art. 417. Q. E. D.

419. COR. 1. If the arcs subtended by the sides of a regular inscribed polygon be bisected, and each point of bisection be joined to the nearest vertices of the polygon, a regular inscribed polygon of double the number of sides is formed.

420. COR. 2. The perimeter of an inscribed polygon is less than the perimeter of an inscribed polygon of double the number of sides.

PROPOSITION II. THEOREM

421. A circle may be circumscribed about, and a circle may be inscribed in, any regular polygon.

To prove that a O may be circumscribed about, or inscribed in, ABCDE.

Proof. I. Through A, B and C (Fig. 1), any three successive vertices of the polygon ABCDE, pass a circumference.

Let O be the center of this circumference.

Art. 235.

Draw the radii OA, OB, OC. Also draw the line OD. Then, in ▲ OBC, OB=OC.

. 40BC= 4 OCB.

(Why ?)

(Why ?)

Hence the circumference which passes through the vertices A, B and C, will also pass through the vertex D. In like manner, it may be proved that this circumference will pass through the vertex E.

Hence a circle described with O as a center, and OA as a radius, will be circumscribed about the given polygon.

II. The sides of the polygon ABCDE (Fig. 2) are equal chords in the circle 0.

Hence they are equidistant from the center. Art. 226. .. a circle described with O as a center, and the distance from 0 to one of the sides of the polygon as a radius, will be inscribed in the given polygon.

Q. E. D.

422. DEF. The center of a regular polygon is the common center of the inscribed and circumscribed circles, as the point in the above figure.

423. DEF. The radius of a regular polygon is the radius of the circumscribed circle, as OA in the above figure.

424. DEF. The apothem of a regular polygon is the radius of the inscribed circle.