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Chapter 4 : Similar Triangles. Informally, similar triangles can be thought of as having the same shape but different sizes. If you had a picture of a triangle and you enlarged it, the result would be similar to the original triangle. Here is the precise definition(Fig.4.1).

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chapter 4 similar triangles
Chapter 4 : Similar Triangles
  • Informally, similar triangles can be thought of as having the same shape but different sizes. If you had a picture of a triangle and you enlarged it, the result would be similar to the original triangle. Here is the precise definition(Fig.4.1)
slide2

Definition. Two triangles and are said to be similar with ration k if the following conditions hold for some positive real number k:

1. 2. 3. 4. DE = k . AB 5. EF = k . BC 6. DF = k . AC

  • The theory of similar triangles is similar in many ways to the theory of congruent triangles. In particular, we have analogues of the three basic congruence theorems ASA, SAS, and SSS. These similarity theorems are the main subject of this section.
slide3

Theorem (ASA for similar triangles). If , , and if DE = k . AB, then and are similar with ratio k.

  • Proof.The proof will be based on comparing the areas of two triangles. Since we know from Section 3.3 that the ratio of the areas is the same as the ratio of the products of the including sides. Hence
  • Since DE = k . AB Likewise, since we get
slide4

We also know that , since , , and the sum of three angles in each triangle is . Hence

  • We now compare these three equations. Comparing (4.1) and (4.3), we yields
  • We can cancel from both sides and then multiply both sides by BC. The result is EF = k . BC. Likewise, if we work with (4.2) and (4.3), we obtain DF = k . AC. This completes the proof.
slide5

The other two basic similarity theorems follow from this one in combination with the appropriate theorems for congruent triangles.

  • Theorem (SAS for similar triangles). If , DE = k . AB and DF = k . AC, then and are similar with ratio k.
slide6

Proof. We construct a third triangle such that , , and . Since DE = k . AB, it follows that . Thus by ASA for similar triangles, and are similar with ratio k. Hence . But DF is also equal to k . AC, which implies . Now compare and : , , , and so these triangles are congruent by SAS. Hence and the theorem now follows by ASA for similar triangles.

  • Theorem (SSS for similar triangles). If DE = k . AB, EF = k . BC, and DF = k . AC, then and are similar with ratio k.
slide7

Proof. Construct a third triangle such that , and . The proof is now parallel to the previous proof: and are similar with ratio k by SAS for similar triangles. This in turn implies that and consequently by SSS.

slide8

In most applications of similar triangles we will use a weaker property than the one we have been studying. We say that and are similar-without specifying k-if for some k the triangles are similar with ratio k. If you look at the definition at the beginning of this chapter, you will see that k can be calculated from the triangles themselves as

  • This leads the following definition.
slide9

Definition. Two triangles and are said to be similar if and if the three numbers ,and are all equal; that is,

  • In this case we write .
  • We may now rewrite our three basic theorems using this new definition, without mention of the ratio k.
  • Theorem (ASA for similar triangles). (also called the AA theorem) If and then .
slide10

Theorem (SAS for similar triangles). If and , , then .

  • Theorem (SSS for similar triangles). If , then .
  • Consider an example. Let AB = 6, BC = 12, and AC =15 and let DE = 60, EF =120, and DF = 150. Then by SSS for similar triangles. We see this by comparing the sides of with the sides of and noting that each side of is 10 times bigger than the corresponding side of (k = 10). However, we cold also compare the sides of to each other and the sides of to each other and compare the ratios:
slide11

BC is twice as long as AB and EF is twice as long as DE; AC is times as long as AB, etc. Algebraically, of course, it is obvious that the equation is equivalent to the equation , and likewise for the other ratios. Hence we may rewrite the last two theorems.

  • Theorem (SAS for similar triangles). If and , then .
  • Theorem (SSS for similar triangles). If and , , then .
applications to constructions
Applications to Constructions
  • In this section we construct geometric solutions to three types of algebraic equations. The first is , the second type is the pair of simultaneous equations
  • And the third type is similar to the second with x – y replacing x + y.
  • Problem. Given line segments of lengths a, b, and c, construct a line segment of length x such that .
slide13

Solution. Let be any line. As in Fig. 4.4, construct points A, B, C on such that AB = b and AC = c. Let be another line through A and construct D such that AD = a. Draw and construct a line through B, parallel to , which intersects at the point X. Then AX = x is the required distance.

slide14

Proof. Since since they are corresponding angles, and . Hence, by the AA theorem, . So , or .

  • Problem. Given a line segment of length a and two other segments with lengths b and c, find a point C in such that .
  • Solution. Construct a line through A (not equal to ) and a line parallel to it through B. Construct points D on the first line and E on the second, as in Fig. 4.5, such that AD = b, BE = c, and such that D and E are on opposite sides of . Then C will be the intersection point of and . Moreover, if AC = x and CB = y, then x and y
slide15

are solutions to the system of equations x + y = a, .

  • Proof. Since as alternate interior angles. Also , by the vertical angle theorem. Hence by AA. So .
slide16

Problem. Given a line segment and two other segments with lengths b and c, with , find a point C on such that .

  • Solution. As in the previous problem, we construct points D and E such that , AD = b, BE = c, but in this case we take D and E to be on the same side of (Fig. 4.6.).
slide17

Proof. As before, the proof follows from .

  • As a further application of these ideas we show how to multiply a line segment by a rational number. For example, say we would like to find a line segment two-fifths as long as given segment . We draw another line through A, as in Fig. 4.7, and choose any point on that line. Then, moving further along the line we construct , and such that . Join points and B and draw a line through parallel to . Then C will be the intersection point of with

By our first construction, .