Exercise 20 Page 566

Let's review the theorems that can help us prove that two triangles are similar.

1. Angle-Angle Similarity Theorem: If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.
2. Side-Side-Side Similarity Theorem: If the corresponding side lengths of two triangles are proportional, then the triangles are similar.
3. Side-Angle-Side Similarity Theorem: If the lengths of two sides of one triangle are proportional to the lengths of two corresponding sides of another triangle and the included angles are congruent, then the triangles are similar.

Now we will identify the similar triangles and find the measures, one at a time.

Similar Triangles

Notice that ∠ C and ∠ B are congruent. We can also see that since both ∠ CFA and ∠ BFD are supplementary angles, they are also both congruent right angles. This means that two angles of △ ACF are congruent to two angles of △ DBF. Therefore, by the Angle-Angle Similarity Theorem, they are similar. △ ACF ~ △ DBF

Finding the Measures

Using our similarity statement from above, we can identify two pairs of corresponding sides that will help us find the requested lengths. AC corresponds with DB CF corresponds with BF Recall that corresponding segments of similar figures will have proportional lengths. We are given expressions for the lengths of these sides which we can use to write a proportion. AC/DB = CF/BF ⇕ 20/2x+1 = 12/2x-1 Let's solve this equation to find x.

20/2x+1 = 12/2x-1

▼

Solve for x

CrossMult

Cross multiply

20(2x-1)=(2x+1)12

Distr

Distribute 20

40x-20=(2x+1)12

Distr

Distribute 12

40x-20=24x+12

SubEqn

LHS-24x=RHS-24x

16x-20=12

AddEqn

LHS+20=RHS+20

16x=32

DivEqn

.LHS /16.=.RHS /16.

x=2

Now that we know the value of x, we can find DB and CB. We will substitute x=2 into the expressions for the lengths.

Measure	Expression	x=2	Simplified
DB	2x+1	2(2)+1	5
CB	12+2x-1	12+2(2)-1	15

We found that DB=5 and CB=15.