an angle CBA equal to the given angle, and take BA equal to the other given side: join AC, and the triangle will evidently be constructed.

PROPOSITION IX. PROBLEM.

A

Two sides of a triangle, and an angle opposite to one of them being given, to construct the triangle.. At the extremity B, of any straight line BC, make an angle CBA equal to the given angle, and make BA equal to that side which is adjacent to the given angle; and from A, as a centre, with a radius equal to the other side, describe an arc, which must either touch, or cut, the line

BC, otherwise a triangle could not be formed. If it touch BC, a line from A to the point of contact D, will be perpendicular to BC (Prop. IX. B. III.), and the right angled triangle ABD will be the triangle required.

The given angle in this case must, therefore, be acute.

But, if instead of touching, the arc cuts BC in two points E, F; then, supposing still that the given angle is acute, if lines be drawn from these points to A, it is obvious that two triangles ABE, ABF, will be formed, each of which will contain the proposed given parts; but will differ in other respects, the angle opposite AB being obtuse in the one, and acute in the other. In order, therefore, to avoid this ambiguity, it is requisite previously to know whether the angle opposite the other given side be acute or obtuse.

If, however, the given angle be obtuse, no ambiguity can arise; for having formed the obtuse angle as BEA, and having made EA equal to the adjacent side, the arc described from A as a centre, with a radius equal to AB, the other side, would cut BC on opposite sides of E (Prop. XXII. Schol. B. I.); so that only one obtuse angled triangle could be formed.

And if the given angle were right, although two triangles would be formed, yet, as the hypothenuses would meet BC at equal distances from the common perpendicular, these triangles would be equal.

PROPOSITION X. PROBLEM.

The three sides of a triangle being given, to construct it.

B

Make BC equal to one of the sides, and from B, as a centre, with one of the other sides as a radius, describe an arc; and from C, as a centre, with a radius equal to the third side, describe another arc, cutting the former in a point A (Prop. XXI. B. I., and Prop. XIII. Schol. 4. B. III.); then, if AB, AC, be drawn, the triangle will be constructed.

Scholium.

From the last three problems it appears that, of the six parts composing a triangle, viz. the sides and the angles, it is necessary to know but three, and their relative positions, in order to determine the triangle. It is, however, requisite that at least one of the given parts be a side, and, moreover, in the case where two sides and an opposite acute angle are the given parts, it is indispensable to know, in order to avoid ambiguity, whether the angle opposite the other given side be acute or obtuse.

PROPOSITION XI. PROBLEM.

Through a given point C, to draw a straight line parallel to a given straight line AB.

To any point P in AB draw a straight line from C, then make the angle PCD equal to the angle APC, and CD will be parallel to AB (Prop. XII. B. I.).

PROPOSITION XII. PROBLEM.

P

Two adjacent sides of a rhomboid, with the angle which they include being given, to construct the rhomboid. Make AB equal to one of the given sides, and the angle A equal to the given angle; then take AD equal to the other given side, and from D, as a centre, with a radius equal to AB, describe an arc; and

Α'

B

from B, as a centre, with a radius equal to AD, describe another; and from C, the point where they intersect, draw CB, CD, and the rhomboid will be completed; for the opposite sides are, by construction, equal (Prop. XXVIII. B. I.).

To make a square equivalent to two given squares. Draw two indefinite lines AB, AC, perpendicular to each other. Take AB equal to the side of one of the given squares, and AC equal to the side of the other: join BC, which will be the side of the proposed square, as is evident from Prop. X. B. II.

Scholium.

B

Any number of squares may be reduced to a single one, by reducing two into one, this and a third into another; then, again, this last and a fourth into another, and so on.

To make a square equivalent to the difference of two given squares.

Draw, as in last problem, the lines AB, AC (see the diagram) perpendicular to each other, making AB equal to the side of the less square; then, from B as a centre, with a radius equal to the side of the other square, describe an arc intersecting AC in C, and AC will be the side of the required square (Prop. X. Cor. 1. B. II.).

PROPOSITION XV. PROBLEM.

To make a rectangle equivalent to a given triangle,

the triangle, and half its base (Prop. III. Cor. 5. B. II.).

To make a triangle equivalent to any given polygon,

ABCDE.

G

D

Draw the diagonal CE, cutting off the triangle CDE; draw DF parallel to CE, meeting AE produced, and join CF: the polygon ABCDE will thus be reduced to the polygon ABCF, having fewer sides by one; for the triangle CDE cut off, is equivalent to the triangle CFE added (Prop. III. Cor. 2. B. II.). Draw, now, the diagonal CA and BG parallel to it, meeting the production of EA: join CG, and the polygon ABCF will be reduced to an equivalent polygon, with fewer sides by one; for the triangle, cut off by CA, has been supplied by the equivalent triangle CGA: and thus, by continually diminishing the sides, the polygon is at length reduced to an equivalent triangle.

Cor. Since a triangle may be converted into an equivalent rectangle, it follows that any polygon may be reduced to an equivalent rectangle.

A rectangle being given, to construct an equivalent one having a side of a given length.

making GH, CK, each equal to BF; then join HK, and the rectangle GK will be equivalent to the rectangle AC, as appears from Proposition I. Book II.

Cor. Hence a polygon may be converted into an equivalent rectangle of a given base, or of a given altitude (Prop. XVI. Cor.).

PROPOSITION XVIII. PROBLEM.

Having given a circumference, or an arc, to find the centre of the circle.

Take any three points A, B, C, in the arc, bisect the distances AB, BC, by the perpendiculars DF, EF (Prop. I.), these perpendiculars will meet in a point F equally distant from the points A, B, C (Prop. VIII. B. III.); F, therefore, is the centre of the circle.

A

B

E

D

F

Scholium.

By a similar construction a circumference may be circumscribed about a given triangle. If the triangle be right angled, the middle of the hypothenuse will be the centre of the circumscribed circle.

To draw a tangent to a circle from a given point A, in the circumference.

Draw the radius CA, and make AB perpendicular to it; AB will be the tangent required (Prop. IX. B. III.).

B

From a given point B, in the arc ABE, of a circle, to draw a tangent thereto, without making use of the

centre.