Exercise 30 Page 589

Let's begin with recalling one of the Special Segments of Similar Triangles Theorems. If two triangles are similar, the lengths of corresponding medians are proportional to the lengths of corresponding sides. Now let's look at the given diagram. Notice that △ ADF is similar to △ ABC by Angle-Angle Similarity Theorem. Let x be the length of BK.

Since we are given that AE and AK are medians of these triangles, we can create a proportion using the fact that the lengths of corresponding medians are proportional to the lengths of corresponding sides. DE/BK=AD/AB Since a median of a triangle divides a side into two congruent segments, the length of DE is also 3 13. Notice that, by the Segment Addition Postulate, the length of AB is the sum of lengths of AD,DG, and GB. 3 13/x=13/13+ 13+ 13 Now we will solve the proportion for x using cross multiplication.

3 13/x=13/13+13+13

AddTerms

Add terms

3 13/x=13/39

▼

Solve for x

ReduceFrac

a/b=.a /13./.b /13.

3 13/x=1/3

CrossMult

Cross multiply

3 13*3=x*1

RearrangeEqn

Rearrange equation

x*1=3 13*3

MultByOne

a * 1=a

x=3 13*3

MixedToFrac

a bc=a* c+b/c

x=10/3*3

FracMultDenomToNumber

a/3* 3 = a

x=10

The length of BK is 10 inches.