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Cases of similar Triangles.

they have the sides of the one respectively parallel to those of the other.

Proof. For, in this case, the angles are equal by § 29. 176. Corollary. The parallel sides are homologous. 177. Theorem. Two triangles are similar, when the sides of the one are equally inclined to those of the other, each to each, as ABC, DEF (fig. 99).

Proof. For if one of the triangles is turned around, by a quantity equal to the angle made by the sides of the one with those of the other, the sides of the two triangles become respectively parallel, and they are, therefore, by § 175, equiangular and similar.

178. Corollary. Two triangles are similar, when the sides of the one are respectively perpendicular to those of the other, and the perpendicular sides are homologous.

179. Theorem. Two triangles ABC, DEF (fig. 98) are similar, if they have an angle A of the one equal to an angle D of the other, and the sides including these angles proportional, that is,

AB: DF AC: DE.

Proof. Place the angle D upon A; E falls upon E', and Fupon F; and E'F is parallel to BC, by § 162, be

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that is, the triangles ABC and DEF are equiangular, and, by 172, similar.

180. Theorem. Two triangles ABC, DEF (fig. 98) are similar, if they have their homologous sides proportional, that is,

Cases of similar Triangles.

AB: DFAC: DE

=

BC: EF.

Proof. Take AE' = DE, and draw EF parallel to BC. The triangles AEF and ABC are similar, by § 175, and we are to prove that AE F is equal to DEF.

Now, by 160,

AE: ACAF' : AB,

and, by hypothesis,

DE or AE: ACDF: AB.

Hence, on account of the common ratio AE' : AC,

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that is, AF and DF are in the same ratio to AB, and are consequently equal.

In the same way it may be proved that EF and EF, being in the same ratio to BC, are equal; and as the triangle DFE has its sides equal to those of AE'F', it is equal to AE'F', and is, therefore, similar to ABC.

181. Theorem. Lines AF, AG, &c. (fig. 100), drawn at pleasure through the vertex of a triangle, divide proportionally the base BC and its parallel DE, so that

DI: BFIK: FG KL: GH, &c.

=

Proof. Since DI is parallel to BF, the triangles ADI, ABF are equiangular, and give the proportion,

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also, since IK is parallel to FG,

Al: AFIK: FG ;

and, therefore, on account of the common ratio Al: AF, DI: BFIK: FG.

It

may

Right Triangle divided into two similar Right Triangles.

be shown in like way, that

IK: FG KL: GH, &c.

=

182. Corollary. When BC is divided into equal parts, the parallel DE is likewise divided into equal parts.

183. Theorem. The perpendicular AD (fig. 101) upon the hypothenuse BC of the right triangle BAC from the vertex A of the right angle, divides the triangle into two triangles BAD, CAD, which are similar to each other and to the whole triangle BAC,

Proof. a. The right triangles BAC and BAD are similar, by $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: BA=BA : 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

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To find a Mean proportional.

drawn to the diameter BC, it is a mean proportional between the segments BD, DC of the diameter.

Proof. For, if the chords AB, AC are drawn, the triangle BAC is, by § 109, right-angled at A.

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

Proof. Join AD and CB. In the triangles AOD and COB, the angles AOD and COB are equal, by § 23; also the angles ADO and CBO are equal, by § 108, because they are each measured by half the arc AC, and, therefore, the triangles AOD and COB are similar by 173, and give the proportion

40: DO= CO: BO.

190. Theorem. If, from a point 0 (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,

40:D0= CO: B0.

To divide a line in extreme and mean ratio.

Proof. Join AC and BD. In the triangles AOC and BOD, the angle O is common, and the angles BAC and BDC are equal, by § 108; these triangles are, therefore, similar, by § 173, and give the proportion

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191. Theorem. If, from a point 0 (fig. 106), taken without a circle, a tangent OC and a secant OA be drawn, the tangent is a mean proportional between the entire secant and the part without the circle, that is,

AO: COCO: BO.

Proof. 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)

40: D0=C0 : BO,

becomes (fig. 106)

AO: COCO: 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

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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.

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

AF: ABAB : AE ;

and, by the theory of proportions,

AB: AF-ABAE: AB-AE.

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