For, the opposite sides are parallel by construction; and consequently, the figure is a parallelogram (D. 28); it is also formed with the given sides and given angle.

PROBLEM XIII.

To find the centre of a given circumference.

Take any three points A, B, and C, on the circumference or arc, and join thein by the chords AB, BC; BC; bisect these chords by the perpendiculars DE and FG then will their point of intersection O. be the centre required (P. VII.).

Scholium. The same construc

A

B

tion enables us to pass a circumference through any three points not in a straight line. If the points are vertices of a triangle, the circle will be circumscribed about it.

PROBLEM XIV.

Through a given point, to draw a tangent to a given circle.

There may be two cases: the given point may lie on the circumference of the given circle, or it may lie without the given circle.

1o. Let C be the centre of the

given circle, and A a point on the circumference, through which the tangent is to be drawn.

Draw the radius CA, and at

A

draw AD perpendicular to AC: then

will AD be the tangent required (P. IX.).

2o. Let C be the centre of the given circle, and A a

point without the circle, through which the tangent is to be drawn.

Draw the line AC; bisect it at

0, and from 0

as a centre, with a radius OC, describe the circumference ABCD; join the point A with the points of intersection D and B: then will both AD and AB be tangent to the given circle, and there will be two solutions.

For, the angles ABC and ADC

are right angles (P. XVIII., C. 2):

hence, each of the lines AB and AD is perpendicular to a radius at its extremity; and consequently, they are tangent to the given circle (P. IX.).

AD,

and

Corollary. The right-angled triangles ABC and ADC, have a common hypothenuse AC, and the side BC equal to DC; and consequently, they are equal in all their parts (B. I., P. XVII.): hence, AB is equal to the angle CAB is equal to the angle CAD. gents are therefore equal, and the line AC bisects the angle between them.

The tan

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

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 ов, describe the arc AMB: then will the segment AMB be the segment required.

half of the arc AKB (P. XXI.); and the inscribed angle 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 POLYGON 8.

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 are

equal, the arcs BFC and DGE are

similar, the sectors BAC and DOE

are similar, and the segments BFC B

and DGE are similar.

E

4. The ALTITUDE OF A TRIANGLE, is the perpendicular distance from the vertex of either an

gle 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 ide, is called the base of the triangle.