To solve for \(x\) and \(y\), we can use the properties of similar triangles. Since \(AG \parallel BF \parallel CE\), the triangles \( \triangle AGB \), \( \triangle BFC \), and \( \triangle CED \) are similar. ### Step 1: Solve for \(x\) Since \( \triangle BFC \sim \triangle CED \), the ratios of corresponding sides are equal: \[ \frac{BF}{FC} = \frac{CD}{ED} \] Substituting the known values: \[ \frac{5}{x} = \frac{6}{4.5} \] Cross-multiply to solve for \(x\): \[ 5 \times 4.5 = 6 \times x \] \[ 22.5 = 6x \] \[ x = \frac{22.5}{6} = 3.75 \] ### Step 2: Solve for \(y\) Since \( \triangle AGB \sim \triangle BFC \), the ratios of corresponding sides are: \[ \frac{AG}{GB} = \frac{BF}{FC} \] Substitute the known values: \[ \frac{y}{9} = \frac{5}{x} \] Now substitute \(x = 3.75\): \[ \frac{y}{9} = \frac{5}{3.75} \] Cross-multiply to solve for \(y\): \[ y \times 3.75 = 5 \times 9 \] \[ 3.75y = 45 \] \[ y = \frac{45}{3.75} = 12 \] ### Solution - \(x = 3.75\) - \(y = 12\)