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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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
Theorem (ASA for similar triangles). If , , and if DE = k . AB, then and are similar with ratio k.
We also know that , since , , and the sum of three angles in each triangle is . Hence
The other two basic similarity theorems follow from this one in combination with the appropriate theorems for congruent triangles.
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.
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.
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
Definition. Two triangles and are said to be similar if and if the three numbers ,and are all equal; that is,
Theorem (SAS for similar triangles). If and , , then .
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.
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.
Proof. Since since they are corresponding angles, and . Hence, by the AA theorem, . So , or .
Problem. Given a line segment and two other segments with lengths b and c, with , find a point C on such that .
By our first construction, .