body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

4. Parallel Lines and Proportional Parts

Continue to next subchapter

Exercise 40 Page 579

By the Converse Triangle Proportionality Theorem, if a line intersects two sides of a triangle and separates the sides into proportional corresponding segments, then the line is parallel to the third side of the triangle.

x=3

Practice makes perfect

We are given △ADF and we want to determine the value of x so that BC is parallel to DF.

By the Converse Triangle Proportionality Theorem, if a line intersects two sides of a triangle and separates the sides into proportional corresponding segments, then the line is parallel to the third side of the triangle. Let's write a proportion using the expressions for the lengths of the segments. AB/BD = AC/CF ⇕ x+5/12=3x+1/15 Let's solve this equation for x.

x+5/12=3x+1/15

▼

Solve for x

LHS * 12=RHS* 12

x+5=12* 3x+1/15

LHS * 15=RHS* 15

15(x+5)=12(3x+1)

Distribute 15

15x+75=12(3x+1)

Distribute 12

15x+75=36x+12

LHS-12=RHS-12

15x+63=36x

LHS-15x=RHS-15x

63=21x

.LHS /21.=.RHS /21.

3=x

Rearrange equation

x=3

We found that if x=3, ABBD is equal to ACCF, and by this BC is parallel to DF.

Subchapter links

Exercises p.577-581