EXERCISES.

1. Draw a circumference of given radius through two given points.

2. Construct an equilateral triangle, having given one of its sides.

3. At a point on a given straight line, construct an angle of 30°.

4. Through a given point without a given line, draw a line forming with the given line an angle of 30°.

5. A line 8 feet long is met at one extremity by a second line, making with it an angle of 30°; find the centre of the circle of which the first line is a chord and the second a tangent.

6. How many degrees in an angle inscribed in an arc of 135°?

7. How many degrees in the angle formed by two secants meeting without the circle and including arcs of 60° and 110°?

8. At one extremity of a chord, which divides the circumference into two arcs of 290° and 70° respectively, a tangent is drawn; how many degrees in each of the angles formed by the tangent and the chord?

9. Show that the sum of the alternate angles of an inscribed hexagon is equal to four right angles.

10. The sides of a triangle are 3, 5, and 7 feet; construct the triangle.

11. Show that the three perpendiculars erected at the middle points of the three sides of a triangle meet in a common point.

12. Construct an isosceles triangle with a given base and a given vertical angle.

13. At a point on a given straight line, construct an angle of 45°

14. Construct an isosceles triangle so so that the base shall be a given line and the vertical angle a right angle. 15. Construct a triangle, having given one angle, one of its including sides, and the difference of the two other sides.

16. From a given point, A, without a circle, draw two tangents, AB and AC, and at any point, D, in the included arc, draw a third tangent and produce it to meet the two others; show that the three tangents form a triangle whose perimeter is constant.

B

H

17. On a straight line 5 feet long, construct a circular segment that shall contain an angle of 30°. 18. Show that parallel tangents to a circle include semi-circumferences between their points of contact.

19. Show that four circles can be drawn tangent to three intersecting straight lines.

BOOK IV.

MEASUREMENT AND RELATION OF POLYGONS.

DEFINITIONS.

1. SIMILAR POLYGONS are polygons which are mutually equiangular, and which have the sides about the equal angles, taken in the same order, proportional.

2. In similar polygons, the parts which are similarly placed in each, are called homologous.

The corresponding angles are homologous angles, the corresponding sides are homologous sides, the corresponding diagonals are homologous diagonals, and so on.

3. SIMILAR ARCS, SECTORS, or SEGMENTS, in different circles, are those which correspond to equal angles at the

centre.

Thus, if the angles A and O are equal, the arcs BFC and DGE are similar, the sectors BAC and DOE are similar, and the segments BFC and DGE are similar.

4. The ALTITUDE OF A TRIANGLE is the perpendicular distance from the vertex of any angle to

the opposite side, or the opposite side pro

duced.

The vertex of the angle from which the

distance is measured, is called the vertex of

the triangle, and the opposite side is called the base of the triangle.

5. The ALTITUDE OF A PARALLELOGRAM is the perpendicular distance between two opposite

sides.

These sides are called bases; one the

upper, and the other, the lower base.

6. The ALTITUDE OF A TRAPEZOID is the perpendicular distance between its parallel sides.

These sides are called bases; one the

upper, and the other, the lower base.

7. The AREA OF A SURFACE is its numerical value expressed in terms of some other surface taken as a unit. The unit adopted is a square described on the linear unit as a side.

PROPOSITION I. THEOREM.

Parallelograms which have equal bases and equal altitudes, are equal.

Let the parallelograms ABCD and EFGH have equal bases and equal altitudes: then the parallelograms are equal.

D H

BE

F

For, let them be so placed that their lower bases shall coincide; then, because they have the same altitude, their upper bases will be in the same line DG, parallel to AB. The triangles DAH and CBG, have the sides AD and BC equal, because they are opposite sides of the parallelogram AC (B. I., P. XXVIII.); the sides AH and BG equal, because they are opposite sides of the parallelogram AG; the angles DAH and CBG equal, because their sides are

parallel and lie in the same direction (B. I., P. XXIV.): hence, the triangles are equal (B. I., P. V.).

If from the quadrilateral ABGD, we take away the triangle DAH, there will remain the parallelogram AG; if from the same quadrilateral ABGD, we take away the triangle CBG, there will remain the parallelogram AC: hence, the parallelogram AC is equal to the parallelogram EG (A. 3); which was to be proved.

A triangle is equal to one half of a parallelogram having an equal base and an equal altitude.

Let the triangle ABC, and the parallelogram ABFD, have equal bases and equal altitudes: then the triangle is equal to one half of the parallelogram.

For, let them be SO placed that that the base of the triangle shall coincide with the lower base of the parallelogram; then, be

D

E F

B

cause they have equal altitudes, the vertex of the triangle will lie in the upper base of the parallelogram, or in the prolongation of that base.

From A, draw AE parallel to BC, forming the parallelogram ABCE. This parallelogram is equal to the parallelogram ABFD, from Proposition I. But the triangle ABC is equal to half of the parallelogram ABCE (B. I., P. XXVIII., C. 1) hence, it is equal to half of the parallelogram ABFD (A. 7); which was to be proved.

Cor. Triangles having equal bases and equal altitudes are equal, for they are halves of equal parallelograms.