# Comparing the Ratio Method with the Parallel Method

### New York State Common Core Math Geometry, Module 2, Lesson 4

Student Outcomes

• Students understand that the Ratio and Parallel Methods produce the same scale drawing and understand the proof of this fact.
• Students relate the equivalence of the Ratio and Parallel Methods to 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.

Comparing the Ratio Method with the Parallel Method

Classwork

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 π > 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.

Opening Exercise

a. Suppose two triangles, β³ π΄π΅πΆ and β³ π΄π΅π·, share the same base π΄π΅ such that points πΆ and π· lie on a line parallel to π΄π΅ . Show that their areas are equal, that is, Area(β³ π΄π΅πΆ) = Area(β³ π΄π΅π·). (Hint: Why are the altitudes of each triangle equal in length?)

b. Suppose two triangles have different-length bases, π΄π΅ and π΄π΅β², that lie on the same line. Furthermore, suppose they both have the same vertex πΆ 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

Discussion

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 βΉ The ratio method,
and
The ratio method βΉ The parallel method.

The first implication above can be stated as the following theorem:

PARALLEL βΉRATIO THEOREM: Given π΄π΅ and point π not on π΄π΅ , construct a scale drawing of π΄π΅ with scale factor π > 0 using the parallel method: Let π΄β² = π·π,π(π΄), and β be the line parallel to β‘π΄π΅ that passes through π΄β². Let π΅β² be the point where ππ΅ intersects β. Then π΅β² is the same point found by the ratio method, that is, π΅β² = π·π,π(π΅)

PROOF: We prove the case when π > 1; the case when 0 < π < 1 is the same but with a different picture. Construct two line segments π΅π΄β² and π΄π΅β² to form two triangles β³π΅π΄π΅β² and β³π΅π΄π΄β², labeled as π1 and π2, respectively, in the picture below.

The areas of these two triangles are equal,

Area(π1) = Area(π2),

by Exercise 1. Why? Label β³ ππ΄π΅ by π0. Then Area(β³ ππ΄β²π΅) = Area(β³ ππ΅β²π΄) because areas add:

Area(β³ππ΄β²π΅) = Area(π0)+ Area(π2)
= Area(π0)+ Area(π1)
= Area(β³ππ΅β²π΄).

Next, we apply Exercise 2 to two sets of triangles: (1) π0 and β³ππ΄β²π΅ and (2) π0 and β³ππ΅β²π΄.

Therefore,
Area(β³ππ΄β²π΅)/Area(π0) = ππ΄β²/ππ΄, and
Area(β³ππ΅β²π΄)/Area(π0) = ππ΅β²/ππ΅.

Since Area(β³ππ΄β²π΅) = Area(β³ ππ΅β²π΄), we can equate the fractions:
ππ΄β²/ππ΄ = ππ΅β²/ππ΅.

Since π is the scale factor used in dilating ππ΄ to ππ΄β², we know that ππ΄β²/ππ΄ = π; therefore, ππ΅β²/ππ΅ = π, or ππ΅β² = π β ππ΅. This last equality implies that π΅β² is the dilation of π΅ from π by scale factor π, which is what we wanted to prove.

Next, we prove the reverse implication to show that both methods are equivalent to each other.

RATIO βΉPARALLEL THEOREM: Given π΄π΅ and point π not on β‘π΄π΅ , construct a scale drawing π΄β²π΅β² of π΄π΅ with scale factor π > 0 using the ratio method (i.e., find π΄ β² = π·π,π(π΄) and π΅ β² = π·π,π(π΅), and draw π΄β²π΅β² ). Then π΅β² 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 π΄β² = π·π,π(π΄), 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 β parallel to π΄π΅ that passes through π΄β² and label the point where β intersects ππ΅ by πΆ. Then π΅β² is the same as the point found using the parallel method if we can show that πΆ = π΅β².

By the parallel βΉ ratio theorem, we know that πΆ = π·π, π(π΅), that is, that πΆ is the point on ππ΅ such that ππΆ = π β ππ΅. But π΅β² is also the point on ππ΅ such that ππ΅β² = π β ππ΅. 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 πΆπ· is said to split the sides of β³ππ΄π΅ proportionally if πΆ is a point on ππ΄, π· is a point on ππ΅, and ππ΄/ππΆ = ππ΅/ππ· (or equivalently, ππΆ/ππ΄ = ππ·/ππ΅). We call line segment πΆπ· 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:

Lesson Summary

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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