# ssrs barcode font 7.4A Selecting Similar Triangles to Prove a Proportion in Objective-C Printer QR Code 2d barcode in Objective-C 7.4A Selecting Similar Triangles to Prove a Proportion

7.4A Selecting Similar Triangles to Prove a Proportion
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In Solved Problem 7.25, it is given that ABCD in a figure like Fig. 7-9 is a parallelogram, and we must AF AE prove that . To prove this proportion, it is necessary to find similar triangles whose sides are BC FB in the proportion. This can be done simply by selecting the triangle whose letters A, E, and F are in the numerators and the triangle whose letters B, C, and F are in the denominators. Hence, we would prove nAEF , nBCF. AE BC Suppose that the proportion to be proved is . In such a case, interchanging the means leads to AF FB AE AF . BC FB The needed triangles can then be selected based on the numerators and the denominators. AE AF Suppose that the proportion to be proved is . Then our method of selecting triangles could not AD FB be used until the term AD were replaced by BC. This is possible, because AD and BC are opposite sides of the parallelogram ABCD and therefore are congruent.
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7.4B Principles of Similar Triangles
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PRINCIPLE 1: PRINCIPLE 2: PRINCIPLE 3:
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Corresponding angles of similar triangles are congruent (by the definition). Corresponding sides of similar triangles are in proportion (by the definition). Two triangles are similar if two angles of one triangle are congruent respectively to two angles of the other.
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Thus in Fig. 7-10, if /A > /A and /B > /B , then nABC , nA B C .
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CHAPTER 7 Similarity
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Fig. 7-10
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PRINCIPLE 4:
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Two triangles are similar if an angle of one triangle is congruent to an angle of the other and the sides including these angles are in proportion.
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a ar b , then nABC , nA B C . br
Thus in Fig. 7-10, if /C > /C and
PRINCIPLE 5:
Two triangles are similar if their corresponding sides are in proportion.
a ar b br c , then nABC , A B C . cr
Thus in Fig. 7-10, if
PRINCIPLE 6:
Two right triangles are similar if an acute angle of one is congruent to an acute angle of the other (corollary of Principle 3). A line parallel to a side of a triangle cuts off a triangle similar to the given triangle.
PRINCIPLE 7:
Thus in Fig. 7-11, if DE i BC, then nADE , nABC.
Fig. 7-11
Fig. 7-12
PRINCIPLE 8: PRINCIPLE 9:
Triangles similar to the same triangle are similar to each other. The altitude to the hypotenuse of a right triangle divides it into two triangles which are similar to the given triangle and to each other. Triangles are similar if their sides are respectively parallel to each other. Triangles are similar if their sides are respectively perpendicular to each other.
Thus in Fig. 7-12, nABC , nACD , nCBD.
PRINCIPLE 10:
Thus in Fig. 7-13, nABC , nA B C .
PRINCIPLE 11:
Thus in Fig. 7-14, nABC , nA B C .
Fig. 7-13
Fig. 7-14
CHAPTER 7 Similarity
SOLVED PROBLEMS
7.17 Applying principle 2 In similar triangles ABC and A B C (Fig. 7-15), find x and y if /A > /A and /B > /B .
Fig. 7-15
Solution
x Since /A > /A and /B > /B , x and y correspond to 32 and 26, respectively. Hence, 32 y 15 so y 191. from which x 24; also 2 26 20 15 , 20
7.18 Applying principle 3 In each part of Fig. 7-16, two pairs of congruent angles can be used to prove the indicated triangles similar. Determine the congruent angles and state the reason they are congruent.
Fig. 7-16
Solutions
(a) /CBD > /BDA and /BCA > /CAD, since alternate interior angles of parallel lines are congruent (BC i AD). Also, /BEC and /AED are congruent vertical angles. (b) /A > /C and /B > /D, since angles inscribed in the same arc are congruent. Also, /AED and /CEB are congruent vertical angles. (c) /ABC > /AED, since each is a supplement of /DEC. /ACB > /ADE, since each is a supplement of /BDE. Also, /A > /A.
7.19 Applying principle 6 In each part of Fig. 7-17, determine the angles that can be used to prove the indicated triangles similar.
CHAPTER 7 Similarity
Fig. 7-17
Solutions
(a) /ACB and /ADC are right angles. /A > /A. (b) /AEC and /BDC are right angles. /B > /ACE, since angles in a triangle opposite congruent sides are congruent. (c) /ACB is a right angle, since it is inscribed in a semicircle. Hence, /AED > /ACB. /D > /B, since angles inscribed in the same arc are congruent.
7.20 Applying principle 4 In each part of Fig. 7-18, determine the pair of congruent angles and the proportion needed to prove the indicated triangles similar.
Fig. 7-18