let fall the perpendiculars OD, OE, OF, on the sides of the triangle these perpendiculars will all be equal.

:

For, in the triangles BOD and BOE, the angles OBE and OBD are equal, by construction; the angles ODB and OEB are equal, because both are right angles; and consequently, the angles BOD and BOE are also equal (B. I., P. XXV., C. 2), and the side OB is common; and therefore, the triangles are equal in all their parts (B. I., P. VI.) hence, OD is equal to OE. In like manner, it may be shown that OD is equal to OF.

From 0 as a centre, with a radius OD, describe a circle, and it will be the circle required. For, each side is perpendicular to a radius at its extremity, and is therefore tangent to the circle.

Corollary. The lines that bisect the three angles of a triangle all meet in one point.

PROBLEM XVI.

On a given line, to construct a segment that shall contain a given angle.

Produce AB towards D; at B construct the angle

DBE equal to the given angle draw BO perpendicular

to BE, and at the middle point G, of AB, draw GO perpendicular to AB; from their point of intersection 0, as a centre, with a radius OB, describe the arc AMB: then will the segment AMB be the segment required.

AMB is measured by half of the same arc : hence, the angle AMB is equal to the angle EBD, and consequently, to the given angle.

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 are those which correspond to equal angles at the centre.

Thus, if the angles A and are

A

4. The ALTITUDE OF A TRIANGLE, is the perpendicular

distance from the vertex of either angle to the opposite side, or the opposite side produced.

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 will the parallelograms be equal.

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.

H C

G H

BE

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 ᎪᏀ ; 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 tribriangle CBG, there will remain the parallelogram AC: hence, the parallelogram AC is equal to the parallelogram EG (A. 3); which was to be proved.

PROPOSITION II. THEOREM.

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 ABFÐ, have equal bases and equal altitudes: then will the triangle be equal to one-half of the parallelogram.

then, because 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.

equal to

From A, draw AE parallel to BC, forming the parallelogram ABCE. This parallelogram will be 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.