To find a Mean proportional.

Demonstration. a. The right triangles BAC and BAD are similar, by art. 174, for the acute angle B is common to them both.

b. In the same way it may be shown, that DAC is similar to BAC, and, therefore, to BAD.

184. Corollary. From the similar triangles BAD, BAC, we have

BD: BABA: BC,

that is, the leg BA is a mean proportional between the hypothenuse BC and the adjacent segment BD.

a. In the same way AC is a mean proportional between BC and DC.

185. Corollary. From the similar triangles BAD, CAD, we have

BD : DA = DA : DC,

or, the perpendicular DA is a mean proportional between the segments BD, DC of the hypothenuse.

186. Theorem. If from a point A (fig. 102), in the circumference of a circle, a perpendicular AD is drawn to the diameter BC, it is a mean proportional between the segments BD, DC of the diam

eter.

Demonstration. For if the chords AB, AC are drawn, the triangle BAC is, by art. 108, right-angled at A.

187. Corollary. The chord BA is a mean proportional between the diameter BC and the adjacent segment BD.

Intersecting Chords.

Likewise, AC is a mean proportional between BC and and DC.

188. Problem. To find a mean proportional between two given lines.

Solution. Draw the indefinite line AB (fig. 103). Take AC equal to one of the given lines, and BC equal to the other. Upon AB as a diameter describe the semicircle ADB. At C erect the perpendicular CD, and CD is, by art. 186, the required mean proportional.

189. Theorem. The parts of two chords which cut each other in a circle are reciprocally proportional, that is (fig. 104), AO: DO CO: BO.

=

Demonstration. Join AD and CB. In the triangles AOD and COB, the angles AOD and COB are equal, by art. 23; also the angles ADO and CBO are equal, by art. 107, because they are each measured by half the arc AC, and, therefore, the triangles AOD and COB are similar, by art. 173, and give the proportion

AO:DO= C0 : B0.

190. Theorem. If, from a point O (fig. 105), taken without a circle, secants OA, OD be drawn, the entire secants AO and DO are reciprocally proportional to the parts BO and CO without the circle, that is,

AO:DO= CO:BO.

Demonstration. Join AC and BD. In the triangles AOC and BOD, the angle O is common, and the angles BAC and BDC are equal, by art. 107; these triangles are, therefore, similar, by art. 173, and give the proportion

40 : DO = COBO.

To divide a Line in Extreme and Mean Ratio.

191. Theorem. If, from a point O (fig. 106), taken without a circle, a tangent OC and a secant OA be drawn, the tangent is a (common proportional between the entire secant and the part without the circle, that is,

AO: COCO: BO.

Demonstration. When, in (fig. 105), the secant OC is turned about the point O until it becomes a tangent, as in (fig. 106), the points C and D must coincide, CO must be equal to DO, and the proportion (fig. 105)

becomes (fig. 106) AO: CO = CO : BO.

192. Problem. To divide a given line AB (fig. 107) at the point C' in extreme and mean ratio, that is, SO that we may have the proportion

AB AC AC: CB.

Solution. At B erect the perpendicular BD equal to half of AB. Join AD, take DE equal to BD, and AC equal to AE, and C is the required point of division.

Demonstration. Describe the semicircumference EBF with the radius DB to meet AD produced in F; and, by the preceding proposition,

hence

AF

and

AE

AC;

- AB= AF— EF= AE = AC, AB-AE-AB-AC- BC;

Similar Polygons composed of similar Triangles.

and the preceding proportion becomes

AB: AC

AC: BC.

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193. Theorem. If two polygons ABCD, &c., A'B'C'D', &c. (fig. 108) are composed of the same number of triangles ABC, ACD, &c., A'B'C', A'C'D', &c. which are similar each to each and similarly disposed, the polygons are similar.

Demonstration. Since the triangles ABC, &c. are similar to A'B'C', &c., their angles must be equal each to each. Hence the angle A of the first polygon, which is the sum of the angles BAC, CAD, &c. is equal to the angle A' of the second polygon, which is the sum of B'A'C', C'A'D', &c. Also B = B',

C = BCA + ACD = B'C'A' + A'C'D' = C', &c.; the polygons are therefore equiangular with respect to each other.

Their homologous sides are, moreover, proportional, for the similar triangles give

Hence, by art. 169, the polygons are similar.

194. Problem. To construct a polygon similar to a given polygon ABCD, &c. (fig. 108) upon a given line A'B', homologous to the side AB.

Solution. Join AC, AD, &c. Draw A'C', A'D', &c., making the angles B'A'C BAC, CA'D CAD, &c.

=

=

Draw B'C', making the angle A'B'C' = ABC, and meeting A'C' at C. Draw C'D', making the angle A'CD' ACD, and meeting A'D' at D'; and so on.

=

To construct a Polygon similar to a given one.

The polygon A'B'C'D', &c. thus constructed, is the required polygon.

Demonstration. For, by art. 173, the successive triangles A'B'C', A'C'D', &c. are similar to ABC, ACD, &c. each to each, and therefore, by the preceding theorem, the polygons are similar.

195. Theorem. If the similar polygons ABCD, &c. A'B'C'D', &c. (fig. 109) have a side AB of the one equal to the homologous side A'B' of the other, the polygons are equal.

Demonstration. The polygons are, by art. 169, equiangular; they are also equilateral, for, by art. 169, the ratio of BC to B'C' is the same as that of AB to A'B', or the ratio of equality; that is, BC= B'C', and, in the same way CD CD', &c.

=

If, then, A'B' is placed upon AB, B'C' will take the direction of BC, because the angle B': C'; and C will fall upon C, because B'C' = BC; and, in the same way it may. be shown, that D' falls upon D, E' upon E, &c.; so that the polygons coincide, and are equal.

196. Theorem. Two similar polygons ABCD, . &c., A'B'C'D', &c. (fig. 108), are composed of the same number of triangles ABC, ACD, &c., A'B'C', A'C'D', &c., which are similar each to each and similarly disposed.

Demonstration. Construct upon A'B' homologous to AB, by art. 194, a polygon similar to ABCD, &c., and it must also be similar to A'B'C'D' &c., and must therefore, by the