whole perimeter FG+GH + &c. of the other figure, as the diameter AL to the diameter FM.

For, draw the two corresponding diagonals, AC, FH, as also the lines BL, GM. Then, since the polygons are similar, they are equiangular, and their like sides have the same ratio (def. 67); therefore the two triangles ABC, FGH, have the angle B = the angle G, and the sides AB, BC, proportional to the two sides FG, GH; consequently these two triangles are equiangular (th. 86), and have the angle ACB = FHG. But the angle ACB = ALB, standing on the same arc AB; and the angle FHG = FMG, standing on the same arc FG; therefore the angle ALB=FMG (ax. 1). And since the angle ABL = FGM, being both right angles, because in a semicircle; therefore the two triangles ABL, FGM, having two angles equal, are equiangular; and consequently their like sides are proportional (th. 84); hence AB: FG :: the diameter AL: the diameter FM.

In like manner, each side BC, CD, &c. has to each side GH, HI, &c. the same ratio of AL to FM; and consequently the sums of them are still in the same ratio, viz. AB + BC + CD, &c. : FG + GH + HI, &c. :: the diam. AL: the diam. FM (th. 72). Q. E. D.

THEOREM XCI.

Similar figures inscribed in circles, are to each other as the squares of the diameters of those circles.

Let ABCDE, FGHIK, be two similar figures, inscribed in the circles whose diameters are AL and FM; then the surface of the polygon ABCDE will be to the surface of the polygon FGHIK, as AL 2 to FM 2.

as

For, the figures being similar, are to each other as the squares of their like sides, AB2 to FG 2 (th. 88). But, by the last theorem, the sides AB, FG, are as the diameters AL, FM; and therefore the squares of the sides AB 2 to FG 2, squares of the diameters AL2 to FM2 (th. 74). Consequently the polygons ABCDE, FGHIK, are also to each other as the squares of the diameters AL2 to FM2 (ax. 1). Q. E. D. [See fig th. xc.]

the

THEOREM XCII.

The circumferences of all circles are to each other as their diameters.

Let D, d, denote the diameters of two circles, and C, c, their circumferences. then will D d :: C: c, or D: C::d : c.

For (by theor. 90), similar polygons inscribed in circles have their perimeters in the same ratio as the diameters of those circles.

Now, as this property belongs to all polygons, whatever the number of the sides may be; conceive the number of the sides to be indefinitely great, and the length of each infinitely small. till they coincide with the circumference of the circle, and be equal to it, indefinitely near. Then the perimeter of the polygon of an indefinite number of sides, is the same thing as the circumfer ence of the circle. Hence it appears that the circumferences of the circles, being the same as the perimeters of such polygons, are to each other in the same ratio as the diameters of the circles. Q. E. D.

THEOREM XCIII.

The areas or spaces of circles, are to each other as the squares of their diameters, or of their radii.

Let A, u, denote the areas or spaces of two circles, and D, d, their diameters; then A:a:: D2: d2.

For (by theorem 91), similar polygons inscribed in circles are to each other as the squares of the diameters of the circles.

Hence, conceiving the number of the sides of the polygons to be increased more or more, or the length of the sides to become less and less, the polygon approaches nearer and nearer to the circle, till at length, by an infinite approach, they coincide, and become in effect equal; and then it follows, that the spaces of the circles, which are the same as of the polygons, will be to each other as the squares of the diameters of the circles. Q. E. D.

Corol. The spaces of circles are also to each other as the squares of the circumferences; since the circumferences are in the same ratio as the diameters (by theorem 92).

The area of any circle, is equal to the rectangle of half its circumference and half its diameter.

Conceive a regular polygon to be inscribed in a circle; and radii drawn to all the angular points, dividing it into as many equal triangles as the polygon has sides, one of which is ABC, of which the altitude is the perpendicular CD from the centre to the base AB.

A

B

Then the triangle ABC, being equal to a rectangle of half the base and equal altitude (th. 26, cor. 2), is equal to the rectangle of the half base AD and the altitude CD; consequently, the whole polygon, or all the triangles added together which compose it, is equal to the rectangle of the common altitude CD, and the halves of all the sides, or the half perimeter of the polygon.

Now, conceive the number of sides of the polygon to be indefinitely increased; then will its perimeter coincide with the circumference of the circle, and consequently the altitude CD will become equal to the radius, and the whole polygon equal to the circle. Consequently, the space of the circle, or of the polygon in that state, is equal to the rectangle of the radius and half the circumference. Q. E. D.

C C

PROBLEMS.

PROBLEM I

To make an equilateral triangle on a given line AB,

From the centres A and B, with the distance AB, describe arcs, intersecting in C. Draw AC, BC, and ABC

will be the equilateral triangle.

For the equal radii AC, BC, are, each of them, equal to AB.

A

B

PROBLEM II.

To bisect a given angle BAC.

From the centre A, with any radius, describe an arc, cutting off the equal lines AD, AE; and from the two centres D, E, with the same radius, describe arcs intersecting in F; then draw AF, which will bisect the angle A as required.

Join DF, EF.

Then the two triangles ADF, AEF,

having the two sides AD, DF, equal to the two AE, EF

(being equal radii), and the side AF common, they are

E

mutually equilateral; consequently, they are also mutually equiangular (th. 5), and have the angle BAF equal to the angle CAF.

Scholium. In the same manner is an arc of a circle bisected.

PROBLEM III.

To bisect a given line AB.

From the two centres A and B, with any equal radii, describe arcs of circles, intersecting each other in C and D; and draw the line CD, which will bisect the given line AB in the point E.

Draw the radii AC, BC, AD, BD. Then, because all these four radii are equal, and the side CD common, the two triangles ACD, BCD, are mutually equilateral; consequently, they are also mutually equiangular (th. 5), and have the angle ACE equal to the angle BCE.

E

Hence, the two triangles ACE, BCE, having the two sides AC, CE, equal to

the two sides BC, CE, and their contained angles equal, are identical (th. 1), and therefore have the side AE equal to EB.

PROBLEM IV.

At a given point C, in a line AB, to erect a perpendicular From the given point C, with any radius, cut off anv equal parts CD, CE, of the given line; and, from the two centres D and E, with any one radius, describe arcs intersecting in F; then join CF, which will be perpendicular as required.

AD C EB

Draw the two equal radii DF, EF. Then the two triangles CDF, CEF, having the two sides CD, DF, equal to the two CE, EF, and CF common, are mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the two adjacent angles at C equal to each other; therefore, the line CF is perpendicular to AB (def. 11).

OTHERWISE.

When the point C is near the end of the line. From any point D, assumed above the line, as a centre, through the given point C describe a circle, cutting the given line at E; and through E and the centre D, draw the diameter EDF; then join CF, which will be the perpendicular required.

For the angle at C, being an angle in a semicircle, is

A E

a right angle, and therefore the line CF is a perpendicular (by def. 15).

CB

PROBLEM V.

From a given point A, to let fall a perpendicular on a given line BC. From the given point A as a centre, with any convenient radius, describe an arc, cutting the given line at the two points D and E; and from the two centres D, E, with any radius, describe two arcs, intersecting at F; then draw AGF, which will be perpendicular to BC as

required.

Draw the equal radii AD, AE, and DF, EF. Then the

two triangles ADF, AEF, having the two sides AD, DF, equal to the two AE, EF, and AF, common, are mutually equilateral; consequently, they are also mutually equiangular (th. 5), and have the angle DAG equal the angle EAG. Hence then, the two triangles ADG, AEG, having the two sides AD, AG, equal to the two AE, AG, and their included angles equal, are therefore equiangular (th. 1), and have the angles at G equal; consequently AG is perpendicular to

BC (def. 11).

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OTHERWISE.

When the point is nearly opposite the end of the line.

From any point D, in the given line BC, as a centre, describe the arc of a circle through the given point A, cutting BC in E; and from the centre E, with the radius EA, describe another arc, cutting the former in F; then draw AGF, which will be perpendicular to BC as required.

B

EC

Draw the equal radii DA, DF, and EA, EF. Then the two triangles DAE, DFE, will be mutually equilateral; consequently, they are also mutually equiangular (th. 5), and have the angles at D equal. Hence, the two triangles DAG, DFG, having the two sides DA, DG, equal to the two DF, DG, and the included angles at D equal, have also the angles at G equal (th. 1); consequently, those angles at G are right angles, and the line AG is perpendicular to DG.

arc.

PROBLEM VI.

To make a triangle with three given lines AB, AC, BC.

With the centre A, and distance AC, describe an With the centre B, and distance BC, describe another arc, cutting the former in C. Draw AC, BC, and ABC will be the triangle required.

For the radii, or sides of the triangle, AC, BC, are equal to the given lines AC, BC, by construction.

A

A

B.

C

Note. If any two of the lines are not together greater than the third, the construction is impossible.

PROBLEM VII.

At a given point A, in a line AB, to make an angle equal to a given angle C.

From the centres A and C, with any one radius, describe the arcs DE, FG. Then, with radius DE, and centre F, describe an arc, cutting FG in G. Through G draw the line AG, and it will form the angle required.

Conceive the equal lines or radii, DE, FG, to be drawn. Then the two triangles CDE, AFG, being

G

E

FB

mutually equilateral, are mutually equiangular (th. 5), and have the angle at A equal to the angle at C

PROBLEM VIII.

Through a given point A, to draw a line parallel to a given line BC.

From the given point A draw a line AD to any point in the given line BC. Then draw the line EAF making the angle at A equal to the angle at D (by prob. 5); so shall EF be parallel to BC as required.

EA

F

DC