those which are adjacent the equal angles ; in triangles they are those which are opposite the equal angles. Thus, if A = D, B= E, and C=F, AB and DE are homologous sides, as also AC and DF, and BC and EF. The corresponding parts of two figures are called homologous whether they be lines or angles.




Proposition 1. Theorem.— Rectangles of equal altitude are proportional to their bases.

There are two cases: 1. Where the bases are commensurable. 2. Where the bases are incommensurable.*

1. Let AB and CD be two rectangles having equal altitudes, AE and CF; then

AB : CD :: EB : FD. Divide the base EB into any number, as 7, equal parts, and let FD contain 4. Through the points of division draw lines parallel to AE and CF. Then AB and CD will be divided into equal (I. 34) rectangles, of which AB will contain 7 and CD 4.

AB : CD :: 7 : 4. But

EB : FD :: 7 : 4; ..(IV. 15), AB : CD :: EB : FD.



*Quantities are commensurable when they exactly contain the same unit; thus, two lines respectively 7 and 4 feet long are commensurable, but two lines respectively V2 and 4 feet long are incommensurable.

2. They are incommensurable.

Apply the smaller rectangle A AC to the larger AB, so that their bases shall be in the same line, and one angle E in each common;

and if it be not true that AB:AC:: EB:EC,

AB : AC :: EB : ED.

[blocks in formation]

suppose that

Lay off the unit of measure, which we take less than CD, on EB; at least one point of division, as F, will fall between C and D; draw FG parallel to AE.

Then, according to Case 1,

AF : AB :: EF : EB.


AB: AC :: EB : ED;

.. (IV. 13), AF : AC :: EF : ED.

But EF is less than ED, therefore AF is less than AC, which is impossible. Therefore, no other line but EC can be a fourth proportional to AB, AC, and EB.

AB : AC :: EB : EC.

Corollary 1.Parallelograms of equal altitude are proportional to their bases.

For any parallelogram is equivalent to a rectangle having the same base and altitude (I. 33).

Corollary 2.- Triangles of equal altitude are proportional to their bases.

For a triangle is half a parallelogram of the same base and altitude (I. 35, Cor.).

Proposition 2. Theorem.-If a straight line be drawn parallel to one of the sides of a triangle, it will cut the other sides, or those sides produced proportionally; and if the sides of a triangle, or the sides produced, be cut proportionally, the straight line which joins the points of section will be parallel to the other side of the triangle.

Let DE be drawn parallel to BC, one of the sides of the triangle ABC; then

BD: DA :: CE : EA.

[ocr errors]
[blocks in formation]

Join BE, CD; then the triangle BDE is equal to the tri- . angle CDE, because they are on the same base DE, and between the same parallels, DE, BC; and ADE is another triangle; and equal quantities have (IV. 14) the same ratio to the same quantity.

BDE: ADE :: CDE : ADE. But (V. 1, Cor. 2)

BDE : ADE :: BD : DA, the common altitude being the perpendicular drawn from E to AB.

Similarly, CDE : ADE :: CE : EA; .. (IV. 15), BD : DA :: CE : EA.

Next, let BD : DA :: CE : EA; then will DE be parallel to BC. Because

BD :

DA CE : EA, and (V.1, Cor. 2), BD: DA :: BDE : ADE; also,

CE : EA :: CDE : ADE; .. (IV. 15), BDE : ADE :: CDE : ADE. Therefore BDE and CDE have the same ratio to ADE; therefore (IV. 14), BDEN CDE.

Hence they are between the same parallels (I. 37); that is, DE is parallel to BC.


Corollary.—(See first figure.)

AD: DB :: AE: EC, by composition, AD+DB : AD :: AE+EC : AE, or,

AB : AD :: AC: AE.

Proposition 3. Theorem.-If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base will have the same ratio, which the other sides of the triangle have to each other.

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD, which meets the base at D; then

BD: DC:: BA : AC. Through C draw CE parallel to AD, meeting BA produced in E. Then, because AC meets the parallel lines AD, CE, the angles ACE, DAC are equal (I. 27). But the angle CAD (Hyp.) is



equal to BAD; therefore ACE is equal to BAD. Again, because BAE meets the parallel lines AD, EC, the angles BAD, AEC are equal; but BAD was shown equal ACE, therefore the angles AEC, ACE are equal, and the side AC is equal (I. 14) to the side AE.

Now, because EC is drawn parallel to AD, (V.2),

BD: DC :: BA : AE. But

AE- AC; .. (IV. 14), BD : DC :: BA : AC.

Proposition 4. X Theorem.The sides about the equal angles of mutually equiangular triangles are proportional; and the homologous sides are the antecedents or consequents of the ratios.

Let ABC, DEF be triangles equiangular to each other, having the angle A equal to the angle D, B equal to E, and C equal to F. The sides about the equal angles will be proportional.

Let the triangle DEF be so placed that the angle D will be at A, and the line DE on the line AB; and because the angle D is equal to the angle A, the line DF will fall on AC; E will be at some point in AB, as G, and F at some point in AC, as H; join GH; and because the angle AGH is equal to the angle ABC, the side GH is parallel (I. 26) to the gide BC

.. (V.2, Cor.), AB : AG :: AC: AH; alternately (IV.7), AB : AC :: AG: AH, or,

AB AC :: DE: DF.

[ocr errors]
« ForrigeFortsett »