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Common Core For Geometry

Worksheets for 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, β³ π΄π΅πΆ and β³ π΄π΅π·, share the same base π΄π΅ such that points πΆ and π· lie on a line parallel to π΄π΅ . Show that their areas are equal, that is, Area(β³ π΄π΅πΆ) = Area(β³ 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΅β², that lie on the same line. Furthermore, suppose they both have the same vertex πΆ 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 βΉ The ratio method,

and

The ratio method βΉ The parallel method.

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

**PARALLEL βΉRATIO 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΄β² = π·π,π(π΄), and β be the line parallel to β‘π΄π΅ that passes through π΄β². Let π΅β² be the point
where ππ΅ intersects β. Then π΅β²
is the same point found by the ratio method, that is,
π΅β² = 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΄β² and π΄π΅β² to form two triangles β³π΅π΄π΅β² and β³π΅π΄π΄β², labeled as π1 and π2,
respectively, in the picture below.

The areas of these two triangles are equal,

Area(π1) = Area(π2),

by Exercise 1. Why? Label β³ ππ΄π΅ by π0. Then Area(β³ ππ΄β²π΅) = Area(β³ ππ΅β²π΄) because areas add:

Area(β³ππ΄β²π΅) = Area(π0)+ Area(π2)

= Area(π0)+ Area(π1)

= Area(β³ππ΅β²π΄).

Next, we apply Exercise 2 to two sets of triangles: (1) π0 and β³ππ΄β²π΅ and (2) π0 and β³ππ΅β²π΄.

Therefore,

Area(β³ππ΄β²π΅)/Area(π0) = ππ΄β²/ππ΄, and

Area(β³ππ΅β²π΄)/Area(π0) = ππ΅β²/ππ΅.

Since Area(β³ππ΄β²π΅) = Area(β³ ππ΅β²π΄), we can equate the fractions:

ππ΄β²/ππ΄ = ππ΅β²/ππ΅.

Since π is the scale factor used in dilating ππ΄ to ππ΄β², we know that ππ΄β²/ππ΄ = π; therefore, ππ΅β²/ππ΅ = π, or ππ΅β² = π β ππ΅. This last equality implies that π΅β² 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 βΉPARALLEL THEOREM**: Given π΄π΅ and point π not on β‘π΄π΅ , construct a scale drawing π΄β²π΅β² of π΄π΅ with scale factor
π > 0 using the ratio method (i.e., find π΄
β² = π·_{π,π}(π΄) and π΅
β² = π·_{π,π}(π΅), and draw π΄β²π΅β²
). Then π΅β² 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΄β² = 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΄β² and label the point where β intersects ππ΅ by πΆ. Then π΅β² is the
same as the point found using the parallel method if we can show that πΆ = π΅β².

By the parallel βΉ ratio theorem, we know that πΆ = π·π, π(π΅), that is, that πΆ is the point on ππ΅ such that ππΆ = π β ππ΅. But π΅β² is also the point on ππ΅ such that ππ΅β² = π β ππ΅. 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 πΆπ· is said to split the sides of β³ππ΄π΅
proportionally if πΆ is a point on Μ
πΜ
Μ
π΄Μ
, π· is a point on Μ
πΜ
Μ
π΅Μ
, and
ππ΄/ππΆ = ππ΅/ππ·
(or equivalently,
ππΆ/ππ΄ = ππ·/ππ΅). We call line segment πΆπ· 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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