Comparing the Ratio Method with the Parallel Method
Today, our goal is to show that the parallel method and the ratio method are equivalent; that is, given a figure in the plane and a scale factor 𝑟 > 0, the scale drawing produced by the parallel method is congruent to the scale drawing produced by the ratio method. We start with two easy exercises about the areas of two triangles whose bases lie on the same line, which helps show that the two methods are equivalent.
a. Suppose two triangles, △ 𝐴𝐵𝐶 and △ 𝐴𝐵𝐷, share the same base 𝐴𝐵 such that points 𝐶 and 𝐷 lie on a line parallel to 𝐴𝐵 . Show that their areas are equal, that is, Area(△ 𝐴𝐵𝐶) = Area(△ 𝐴𝐵𝐷). (Hint: Why are the altitudes of each triangle equal in length?)
b. Suppose two triangles have different-length bases, 𝐴𝐵 and 𝐴𝐵′, that lie on the same line. Furthermore, suppose they both have the same vertex 𝐶 opposite these bases. Show that the value of the ratio of their areas is equal to the value of the ratio of the lengths of their bases, that is
To show that the parallel and ratio methods are equivalent, we need only look at one of the simplest versions of a scale drawing: scaling segments. First, we need to show that the scale drawing of a segment generated by the parallel method is the same segment that the ratio method would have generated and vice versa. That is,
The parallel method ⟹ The ratio method,
The ratio method ⟹ The parallel method.
The first implication above can be stated as the following theorem:
PARALLEL ⟹RATIO THEOREM: Given 𝐴𝐵 and point 𝑂 not on 𝐴𝐵 , construct a scale drawing of 𝐴𝐵 with scale factor 𝑟 > 0 using the parallel method: Let 𝐴′ = 𝐷𝑂,𝑟(𝐴), and ℓ be the line parallel to ⃡𝐴𝐵 that passes through 𝐴′. Let 𝐵′ be the point where 𝑂𝐵 intersects ℓ. Then 𝐵′ is the same point found by the ratio method, that is, 𝐵′ = 𝐷𝑂,𝑟(𝐵)
PROOF: We prove the case when 𝑟 > 1; the case when 0 < 𝑟 < 1 is the same but with a different picture. Construct two line segments 𝐵𝐴′ and 𝐴𝐵′ to form two triangles △𝐵𝐴𝐵′ and △𝐵𝐴𝐴′, labeled as 𝑇1 and 𝑇2, respectively, in the picture below.
The areas of these two triangles are equal,
Area(𝑇1) = Area(𝑇2),
by Exercise 1. Why? Label △ 𝑂𝐴𝐵 by 𝑇0. Then Area(△ 𝑂𝐴′𝐵) = Area(△ 𝑂𝐵′𝐴) because areas add:
Area(△𝑂𝐴′𝐵) = Area(𝑇0)+ Area(𝑇2)
= Area(𝑇0)+ Area(𝑇1)
Next, we apply Exercise 2 to two sets of triangles: (1) 𝑇0 and △𝑂𝐴′𝐵 and (2) 𝑇0 and △𝑂𝐵′𝐴.
Area(△𝑂𝐴′𝐵)/Area(𝑇0) = 𝑂𝐴′/𝑂𝐴, and
Area(△𝑂𝐵′𝐴)/Area(𝑇0) = 𝑂𝐵′/𝑂𝐵.
Since Area(△𝑂𝐴′𝐵) = Area(△ 𝑂𝐵′𝐴), we can equate the fractions:
𝑂𝐴′/𝑂𝐴 = 𝑂𝐵′/𝑂𝐵.
Since 𝑟 is the scale factor used in dilating 𝑂𝐴 to 𝑂𝐴′, we know that 𝑂𝐴′/𝑂𝐴 = 𝑟; therefore, 𝑂𝐵′/𝑂𝐵 = 𝑟, or 𝑂𝐵′ = 𝑟 ⋅ 𝑂𝐵. This last equality implies that 𝐵′ is the dilation of 𝐵 from 𝑂 by scale factor 𝑟, which is what we wanted to prove.
Next, we prove the reverse implication to show that both methods are equivalent to each other.
RATIO ⟹PARALLEL THEOREM: Given 𝐴𝐵 and point 𝑂 not on ⃡𝐴𝐵 , construct a scale drawing 𝐴′𝐵′ of 𝐴𝐵 with scale factor 𝑟 > 0 using the ratio method (i.e., find 𝐴 ′ = 𝐷𝑂,𝑟(𝐴) and 𝐵 ′ = 𝐷𝑂,𝑟(𝐵), and draw 𝐴′𝐵′ ). Then 𝐵′ is the same as the point found using the parallel method.
PROOF: Since both the ratio method and the parallel method start with the same first step of setting 𝐴′ = 𝐷𝑂,𝑟(𝐴), the only difference between the two methods is in how the second point is found. If we use the parallel method, we construct the line ℓ parallel to 𝐴𝐵 that passes through 𝐴′ and label the point where ℓ intersects 𝑂𝐵 by 𝐶. Then 𝐵′ is the same as the point found using the parallel method if we can show that 𝐶 = 𝐵′.
By the parallel ⟹ ratio theorem, we know that 𝐶 = 𝐷𝑂, 𝑟(𝐵), that is, that 𝐶 is the point on 𝑂𝐵 such that 𝑂𝐶 = 𝑟 ⋅ 𝑂𝐵. But 𝐵′ is also the point on 𝑂𝐵 such that 𝑂𝐵′ = 𝑟 ⋅ 𝑂𝐵. Hence, they must be the same point. The fact that the ratio and parallel methods are equivalent is often stated as the triangle side splitter theorem. To understand the triangle side splitter theorem, we need a definition:
SIDE SPLITTER: A line segment 𝐶𝐷 is said to split the sides of △𝑂𝐴𝐵 proportionally if 𝐶 is a point on ̅𝑂̅̅𝐴̅, 𝐷 is a point on ̅𝑂̅̅𝐵̅, and 𝑂𝐴/𝑂𝐶 = 𝑂𝐵/𝑂𝐷 (or equivalently, 𝑂𝐶/𝑂𝐴 = 𝑂𝐷/𝑂𝐵). We call line segment 𝐶𝐷 a side splitter.
TRIANGLE SIDE SPLITTER THEOREM: A line segment splits two sides of a triangle proportionally if and only if it is parallel to the third side. Restatement of the triangle side splitter theorem:
THE TRIANGLE SIDE SPLITTER THEOREM: A line segment splits two sides of a triangle proportionally if and only if it is parallel to the third side.
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