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512. If two triangles have two sides of the one proportional to two sides of the other, and an angle in each opposite one corresponding pair of these sides equal, the angles opposite the other pair are either equal or supplemental.

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The angles included by the proportional sides are either equal or unequal. CASE I. If they are equal, then the third angles are equal.

(174. The sum of the angles of a triangle is a straight angle.)

CASE II. If the angles included by the proportional sides are unequal, one must be the greater.

HYPOTHESIS. ABC and FGH As with AB : BC :: FG : GH, * A = * F, X B > * G.

CONCLUSION. C + H = st. 4.
PROOF. Make 4 ABD = XG;

(508. Equiangular triangles are similar.)


:: AB : BD :: FG : GH; therefore, from our hypothesis,

AB : BD :: AB : BC; therefore, by 491, BD = BC,

x C = ABDC. But % BDC + 4BDA = st. 4,

XC + H = st. .


513. COROLLARY. If two triangles have two sides of the one proportional to two sides of the other, and an angle in each opposite one corresponding pair of these sides equal, then if one of the angles opposite the other pair is right, or if they are oblique, but not supplemental, or if the side opposite the given angle in each triangle is not less than the other proportional side, the triangles are similar.

514. In similar figures, sides between equal angles are called Homologous, or corresponding.

The ratio of a side of one polygon to its homologous side in a similar polygon is called the Ratio of Similitude of the polygons. Similar figures are said to be similarly placed when each side of the one is parallel to the corresponding side of the other.


515. If two unequal similar figures are similarly placed, all lines joining a vertex of one to the corresponding vertex of the other are concurrent.

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HYPOTHESIS. ABCD and A'B'C'D two similar figures similarly placed.

CONCLUSION. The lines AA', BB', CC', etc., meet in a point P. PROOF. Since AB and A'

B' are unequal,

AA and BB' are not ll. Call their point of intersection P. Then

AP : A'P :: BP : B'P :: AB : AB'.

(508. Equiangular triangles are similar.)

In the same way, if BB' and CC' meet in Q, then

BQ : B'e :: BC: B'C'. But, by hypothesis,

AB : AB :: BC : B'C',

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(502. A line can be cut only at a single point into external segments having a given


516. COROLLARY. Similar polygons may be divided into the same number of triangles similar and similarly placed.

For if, with their corresponding sides parallel, one of the polygons were placed inside the other, the lines joining corresponding vertices would so divide them.

517. The point of concurrence of the lines joining the equal angles of two similar and similarly placed figures is called the Center of Similitude of the two figures.

518. COROLLARY. The sects from the center of similitude along any line to the points where it meets corresponding sides of the similar figures are in the ratio of those sides.

EXERCISES. 87. Construct a polygon similar to a given polygon, the ratio of similitude of the two polygons being given.


519. A perpendicular from the right angle to the hypothenuse divides a right-angled triangle into two other triangles similar to the whole and to one another.

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PROOF. X CAD = X BAC, and rt. % ADC = rt. 4 ACB,

* ACD = % ABC,
(174. The three angles of any triangle are equal to a straight angle.)

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In the same way we may prove A CBD ~ A ABC;


520. COROLLARY I. Each side of the right triangle is a mean proportional between the hypothenuse and its adjacent segment.

For since A ACD ~ A ABC,


AB : AC :: AC : AD.

521. COROLLARY II. The perpendicular is a mean proportional between the segments of the hypothenuse.

522. COROLLARY III. Since, by 383, lines from any point in a circle to the ends of a diameter form a right angle, therefore, if from any point of a circle a perpendicular be dropped upon a diameter, it will be a mean proportional between the segments of the diameter.


523. The bisector of an interior or exterior angle of a triangle divides the opposite side internally or externally in the ratio of the other two sides of the triangle.

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HYPOTHESIS. ABC any A. BD the bisector of 4 at B.
PROOF. Draw AF || BD.

Then, of the two angles at B given equal by hypothesis, one equals the corresponding interior angle at F, and the other the corresponding alternate angle at A, (168. A line cutting two parallels makes alternate angles equal, and (169) corre

sponding angles equal.)

.:. AB = BF. (126. If two angles of a triangle are equal, the sides opposite are equal.) But

BF :BC :: AD : DC, (499. If a line be parallel to a side of a triangle, it cuts the other sides propor

tionally.) ... AB :BC :: AD: DC.

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