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Fig. 110.

Fig. 111.

190. Corollary IV.

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BC: AC:: BC: CD.

The rectangles BDEF, DCGE, having also the same altitude DE, are to each other as their bases BD, CD. Now these rectangles are equivalent to the squares AH, AI, therefore,

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or, the squares of the two sides of a right angle are to each other, as the segments of the hypothenuse adjacent to these sides.

THEOREM.

191. In a triangle ABC (fig. 110), if the angle C be acute, the square of the side opposite to it will be less than the sum of the squares of the sides containing it, and, AD being drawn perpendicu lar to BC, the difference will be equal to double the rectangle BC-x CD, or,

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Demonstration. The proposition admits of two cases. 1. If the perpendicular fall within the triangle ABC, we shall have BD = BC — CD, and consequently (182)

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BD=BC+ CD-2BC × CD ;

adding AD to each member, we have

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AD+BD=BC + CD + AD — 2BC × CD ;

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2

but the right-angled triangles ABD, ADC, give AD + BD = AB,

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CD+AD=AC; therefore

AB=BC+AC — 2BC × CD.

2. If the perpendicular AD fall without the triangle ABC, we shall have BD = CD- BC, and consequently (182)

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add to each AD, and we shall obtain, as before,

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192. In a triangle ABC (fig. 111), if the angle C be obtuse, the square of the side opposite to it will be greater than the sum of the

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squares of the sides containing it, and, AD being drawn perpendicular to BC produced, the difference will be equal to double the rectangle BC x CD, or,

AB=AC+BC+2BC × CD.

Demonstration. The perpendicular cannot fall within the triangle; for if it should fall, for example, upon E, the triangle ACE would have at the same time a right angle E and an obtuse angle C, which is impossible (75); consequently it falls without, and we have BD=BC + CD, and from this (180)

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Adding to each term AD, and making the reductions as in the preceding theorem, we obtain

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193. Scholium. The right-angled triangle is the which the sum of the squares of two of the sides is equal to the square of the third; for, if the angle contained by their sides be acute, the sum of their squares will be greater than the square of the side opposite; if it be obtuse, the reverse will be true.

THEOREM.

194. In any triangle ABC (fig. 112), if we draw from the vertex Fig. 112. to the middle of the base the line AE, we shall have

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Demonstration. Let fall the perpendicular AD upon the base BC, the triangle AEC will give (191),

AC=AE+EC-2EC × ED;

the triangle ABE will give (192),

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therefore, by adding the corresponding members, and observing that EB = EC, we shall have

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195. Corollary. In every parallelogram the sum of the squares of the sides is equal to the sum of the squares of the diagonals.

For the diagonals AC, BD (fig. 113), mutually bisect each Fig. 113. other in the point E (88), and the triangle ABC gives

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AD + DC = 2AE + 2DE ;

adding the corresponding members and observing that BE = DE, we have

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AB+AD+DC+BC = 4AE + 4DE.

But 4AE is the square of 2AE or of AC; and 4DE is the square of BD; therefore the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals.

Fig. 114.

Fig. 115.

THEOREM.

196. The line DE (fig. 114), drawn parallel to the base of a triangle ABC, divides the sides AB, AC, proportionally; so that AD: DB:: AE: EC.

Demonstration. Join BE and DC; the two triangles BDE, DEC, have the same base DE; they have also the same altitude, since the vertices B and C are situated in a parallel to the base; therefore the triangles are equivalent (170).

The triangles ADE, BDE, of which the common vertex is E, have the same altitude, and are to each other as their bases AD, DB (177); thus,

ADE: BDE :: AD: DB.

The triangles ADE, DEC, of which the common vertex is D, have also the same altitude, and are to each other as their bases AE, EC; that is, ADE: DEC :: AE : EC.

But it has been shown that the triangle BDE = DEC; therefore, on account of the common ratio in the two proportions (111), AD: DB:: AE: EC.

197. Corollary 1.

position (IV)

or

also

198.

We obtain from the above theorem by com

AD+DB: AD::AE+EC : AE,

AB: AD: AC: AE,

AB: BD:: AC: CE.

Corollary 11. If between two straight lines AB, CD,

(fig. 115), parallels AC, EF, GH, BD, &c., be drawn, these two straight lines will be cut proportionally, and we shall have,

AE: CF EG: FH:: GB: HD.

For, let O be the point of meeting of the straight lines, AB,
CD; in the triangle OEF, the line AC being drawn parallel to
the base EF, OE: AE:: OF: CF, or OE: OF:: AE: CF.
In the triangle OGH we have likewise

OE: EG:: OF: FH, or OE: OF :: EG: FH; therefore, on account of the common ratio OE: OF, these two proportions give

AE: CF:: EG: FH.

It may be demonstrated, in the same manner, that

EG: FH:: GB : HD,

and so on; therefore the lines AB, CD, are cut proportionally by the parallels EF, GH, &c.

THEOREM.

199. Reciprocally, if the sides AB, AC (fig. 116), are cut pro- Fig. 116. portionally by the line DE, so that AD: DB:: AE: EC, the line DE will be parallel to the base BC.

Demonstration. If DE is not parallel to BC let us suppose that DO is parallel to it; then, according to the preceding theoAD: DB:: AO: OC.

rem,

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which is impossible, since of the antecedents AE is greater than AO, and of the consequents EC is less than OC; hence the line, drawn through the point D parallel to BC, does not differ from DE; therefore DE is this line.

200. Scholium. The same conclusion might be deduced from the proportion AB: AD:: AC: AE.

For this proportion would give (IV)

AB-AD: AD:: AC-AE: AE, or BD: AD:: EC: AE.

THEOREM.

201. The line AD (fig. 117), which bisects the angle BAC of a Fig. 117. triangle, divides the base BC into two segments BD, DC, proportional to the adjacent sides AB, AC; so that, BD: DC:: AB: AC. Demonstration. Through the point C draw CE parallel to AD to meet BA produced.

In the triangle BCE, the line AD being parallel to the base.
196), GE,
BD: DC:: AB: AE.

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Fig. 119.

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But the triangle ACE is isosceles; for, on account of the parallels AD, CE, the angle ACE DAC, and the angle AEC= BAD (67); and, by hypothesis, DAC = BAD; therefore the angle ACE = AEC, and consequently AE = AC (48); substituting then AC for AE in the preceding proportion, we have

BD: DC:: AB: AC.

THEOREM.

202. Two equiangular triangles have their homologous sides proportional and are similar.

Demonstration. Let ABC, CDE (fig. 119), be two triangles, which have their angles equal, each to each, namely, BAC=CDE, ABC = DCE, and ACB= DEC; the homologous sides, or those adjacent to the equal angles, will be proportional, that is,

BC: CE:: BA: CD:: AC: DE.

Let the homologous sides BC, CE, be in the same straight line, and produce the sides BA, ED, till they meet in F.

Since BCE is a straight line, and the angle BCA = CED, it follows that AC is parallel to DE (67). Also, since the angle ABC DCE, the line AB is parallel to DC; therefore the figure ACDF is a parallelogram.

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In the triangle BFE, the line AC being parallel to the base FE, BC: CE :: BA: AF (196); substituting in the place of AF its equal CD, we have

BC: CE:: BA: CD.

In the same triangle BFE, BF being considered as the base, since CD is parallel to BF, BC: CE::FD:DE. Substituting for FD its equal AC, we have

BC: CE:: AC: DE.

From these two proportions, which contain the same ratio BC: CE, we have

AC: DE:: BA: CD.

Hence the equiangular triangles BAC, CDE, have the homologous sides proportional. But two figures are similar, when they have, at the same time, their angles equal, each to each, and the homologous sides proportional (162); therefore the equiangular triangles BAC, CDE, are two similar figures.

203. Corollary. In order to be similar, it is sufficient that two triangles have two angles of the one respectively equal to two angles of the other; for then the third angles will be equal and the two triangles will be equiangular.

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