Question

splitting triangle sides with dilation (part 2)

1. segment ( ab ) is parallel to segment ( ab ).

a. what is the length of segment ( ab )?
b. what is the length of segment ( bb )?

Explanation:

Step1: Use the basic proportionality theorem (Thales' theorem)

Since \(A'B'\parallel AB\), we have \(\frac{A'B'}{AB}=\frac{CA'}{CA}\). Here \(CA' = 6\), \(CA=6 + 3=9\), and \(A'B' = 5\). Let \(AB=x\). Then \(\frac{5}{x}=\frac{6}{9}\).

Step2: Solve for \(AB\)

Cross - multiply: \(6x=5\times9\). So \(6x = 45\), and \(x=\frac{45}{6}=\frac{15}{2}=7.5\).

Step3: Use the ratio for \(B'B\)

Let \(B'B = y\). We know that \(\frac{B'B}{BB' + B'C}=\frac{3}{9}\) (from the similar - triangle ratio). Also, \(BB' + B'C=7 + y\). But using another ratio from similar triangles \(\frac{B'B}{7 + y}=\frac{3}{9}\). Cross - multiply: \(9y=3\times(7 + y)\). Expand: \(9y=21+3y\). Subtract \(3y\) from both sides: \(9y-3y=21\), \(6y = 21\), \(y=\frac{21}{6}=3.5\)

Answer:

a. The length of segment \(AB\) is \(7.5\).
b. The length of segment \(B'B\) is \(3.5\).