Hence (131) the triangle ABC is to the triangle FGH as the sum of the triangles ABC, ACD, ADE, to the sum of the triangles FGH, FHI, FIK; but the sum of the triangles ABC, ACD, ADE, compose the whole polygon ABCDE, and the sum of the triangles FGH, FHI, FIK, compose the polygon FGHIK; wherefore the triangle ABC is to the triangle FGH as the polygon ABCDE is to the polygon FGHIK; but the triangle ABC is to the triangle FGH as AB' is to FG2; therefore the similar polygons are as the squares of their homologous sides.

166. COROLLARY.-If three similar figures have their homologous sides equal to the three sides of a right-angled triangle, the figure made on the side opposite to the right angle shall be equal to the other two.

THEOREM 68.

167. In any triangle, ABC, the square of AB, opposite to one of the acute angles, is equal to the difference between the sum of the squares of the other two sides, and twice the rectangle BD x DC, made by the perpendicular AD, to the side BC.

There are two cases, according as the perpendicular falls within or without the triangle. In the first case, BD = BC-CD; and, in the second case, BDCD-BC.

In either case...

...

D

ABAD2 + BD9

AD3+ CD2=AC2.

But (160)

and (160)

Therefore, by addition,

168. In any obtuse-angled triangle, the square of the side opposite to the obtuse angle is equal to the sum of the squares of the other two sides, and twice the rectangle, BC x CD, made by the perpendicular, AD, upon the

169. If any two chords, in a circle, cut each other, the rectangle of the segments of the one is equal to the rectangle of the segments of the other. Let AB and CD cut each other in O, then OA × OB = OD × OC.

For, join AC and BD: then, in the triangles AOC, BOD, the vertical angles at O are equal; also the angle AD and C=B (92), consequently the triangles AOC and DOB are similar, and their homologous sides proportional.

Whence AO: OC: DO: OB

Wherefore AO × OB = OD × CO.

THEOREM 71.

170. If any two chords, in a circle, be produced to meet each other, the rectangle of the two distances, from the point of intersection to each extremity of the one chord, is equal to the rectangle of the two distances from the point of intersection to each extremity of the other chord.

Let AB and CD be two chords, and let them be produced to meet in O; OA × OB=OD × OC.

For, join AC and BD; then, in the triangles AOC and DOB the angle at O is common, and the angle A =D (92); therefore the third angle, ACB, of the one triangle, is equal to the third angle, DBO, of the other; consequently the triangles AOC and DOB are similar, and their homologous sides proportional.

Whence AO: OC :: DO: OB

Wherefore, AO × OB = OD × OC.

B

THEOREM 72.

171. In the same circle, or in equal circles, any angles, ACB, DEF, at the centres, are to each other as the arcs AB, DF, of the circles intercepted between the lines which contain the angles.

Let us suppose that the arc AB contains

9

three of such parts as DF contains four. Let Ap, pq, qB, be the equal parts in AB, and Dr, B rs, &c. the equal parts in DF: draw the lines Cp, Cq, Er, Es, &c.; the angles ACp, pCq, qCB, DEr, &c. are all equal; therefore, as the arc AB contains 4th part of the arc DF three times, the angle ACB will evidently contain of the angle DEF also three times; and, in general, whatever number of times the arc AB contains some part of the arc DF, the same number of times will the angle ACB contain a like part of the angle DEF.

THEOREM 73.

172. In two different circles, similar arcs are as the radii of the circles. Let the circles AFG, afg, be each described from the centre C. Draw the radii CA, CF, then the arcs Af and af are similar. Draw CB, indefinitely, near CA, and the sectors Cab, CAB, will approach very nearly to isosceles triangles, which are similar to each other; therefore,

Ca: CA ab: AB; let BF be divided into small arcs, each equal to AB, and draw the radii from each point of division; then bf will contain as many arcs, each equal to ab, as the arc BF contains arcs equal to AB; therefore af is the same multiple of ab that AF is of AB; whence Ca: CA :: af: AF.

THEOREM 74.

173. The circumference of circles are to one another as their diameters.

8.

For, let the circumferences ABCD, abcd, be divided into quadrants by the radii OaA, ObB, OcC, OdD,

then the quadrants AB, ab, will be similar arcs;

therefore, OA: Oa :: AB : ab

wherefore, OA: Oa:: 4AB: 4ab.

a

A

PRACTICAL GEOMETRY.

B

PROBLEM 1.

174. To make an angle at a given point, E, (fig. 35, pl. I,) in a straight line, DE, equal to a given angle ABC.

From the centre B, with any radius, describe an arc gh, cutting BA at g, and BC at h; from the point E, with the same radius, describe an arc, ik, cutting ED ati: make ik equal to gh, and through the point k draw EF: then the angle DEF will be equal to the given angle ABC.

PROBLEM 2.

175. To bisect a given angle ABC (fig. 36, pl. I).

From BA and BC cut off Be and Bf, equal to each other; from the points e and f, as centres, with any radius greater than the distance ef, describe arcs, cutting each other at G, and join BG, which will bisect the angle ABC, as required.

PROBLEM 3.

176. Through a given point g, (fig. 37. pl. I,) to draw a straight line parallel to a given straight line, AB.

From g draw ge, to cut AB at any angle in the point e: in AB take any other point f; make the angle Bfh equal to feg, and make fh equal to

eg, and through the points g and h draw the line CD; then CD will pass through g parallel to AB, as required.

PROBLEM 4.

177. At a given distance, parallel to a given straight line, AB, (fig. 38, pl. I,) to draw a straight line, CD.

In the given straight line AB, take any two points, e and ƒ; and, from the centres e and f, with the given distance, describe arcs at p and q; draw the line CD, to touch the arcs p and q ; then CD will be parallel to AB, at the distance required.

PROBLEM 5.

178. To bisect a given straight line, CD, (fig. 39, pl. I,) by a perpendicular. From the points C and D, with any distance greater than the half of CD, describe arcs cutting each other in A and B : join AB, and this line will bisect AB perpendicularly.

PROBLEM 6.

179. From a given point C, (fig. 40, pl. I,) in a given straight line, AB, to erect a perpendicular.

In the straight line, AB, take any two points, e and f, equally distant from C: from the points e and f, with any equal radius, greater than the half of ef, describe arcs cutting each other at D, and draw CD, which will be perpendicular to AB.

PROBLEM 7.

180. From a given point, B, (fig. 1, pl. II,) at the extremity of a given straight line, AB, to draw a perpendicular.

Take any point, E, above the line AB, and, with the radius BE, describe the arc d BC, cutting AB in d: draw the straight line d EC, and join BC, which will be the perpendicular required.

PROBLEM 8.

181. From a given point C, (fig. 2, pl. II,) to let fall a perpendicular to a