Question

1. complete the following proportions using the figure below.

\\(\frac{b}{c}=\frac{\quad}{f}\\)
\\(\frac{d}{f}=\frac{\quad}{c}\\)
\\(\frac{c}{a + b}=\frac{f}{\quad}\\)

Explanation:

Step1: Analyze Proportional Segments

The figure has parallel lines cut by transversals, so by the Basic Proportionality Theorem (Thales' theorem) or similar triangles, corresponding segments are proportional. For \(\frac{b}{c}=\frac{\_}{f}\), find the corresponding segment to \(b\) on the other transversal. The segment corresponding to \(b\) is \(e\)? Wait, no, looking at the transversals: the first transversal has segments \(a, b, c\) (wait, no, the left transversal: from top, segments \(a\) (between first and second parallel), \(b\) (second and third), \(c\) (third and fourth). The right transversal: \(d\) (first and second), \(e\) (second and third), \(f\) (third and fourth). Wait, maybe I mislabeled. Wait, the left transversal: segments \(a\) (top to second), \(b\) (second to third), \(c\) (third to bottom). Right transversal: \(d\) (top to second), \(e\) (second to third), \(f\) (third to bottom). So since the lines are parallel, the ratios of corresponding segments on the transversals are equal. So \(\frac{b}{c}=\frac{e}{f}\)? Wait, no, the first proportion: \(\frac{b}{c}=\frac{\_}{f}\). Wait, maybe the segments are \(b\) and \(c\) on one transversal, and the other transversal has segments corresponding to \(b\) (which is \(e\)?) No, wait the problem's first proportion: \(\frac{b}{c}=\frac{\_}{f}\). Let's re-express. Wait, maybe the left transversal segments: \(a\) (top - second), \(b\) (second - third), \(c\) (third - bottom). Right transversal: \(d\) (top - second), \(e\) (second - third), \(f\) (third - bottom). So by proportionality, \(\frac{b}{c}=\frac{e}{f}\)? Wait, no, the second proportion: \(\frac{d}{f}=\frac{\_}{c}\). Let's check the third proportion: \(\frac{c}{a + b}=\frac{f}{\_}\). Wait, \(a + b\) is the sum of \(a\) and \(b\), which is the segment from top to third parallel on the left transversal. The corresponding segment on the right transversal is \(d + e\)? Wait, no, maybe I made a mistake. Wait, let's look at the given proportions. First proportion: \(\frac{b}{c}=\frac{\_}{f}\). The segments \(b\) and \(c\) are on the left transversal (between second - third and third - bottom). The right transversal has \(e\) (second - third) and \(f\) (third - bottom). Wait, no, \(b\) is between second and third parallel (left transversal), \(c\) between third and fourth. On the right transversal, between second and third is \(e\), third and fourth is \(f\). So \(\frac{b}{c}=\frac{e}{f}\)? But wait, the second proportion: \(\frac{d}{f}=\frac{\_}{c}\). \(d\) is between first and second (right transversal), \(f\) between third and fourth. On left transversal, between first and second is \(a\), third and fourth is \(c\). Wait, no, maybe the segments are: left transversal: \(a\) (top - second), \(b\) (second - third), \(c\) (third - bottom). Right transversal: \(d\) (top - second), \(e\) (second - third), \(f\) (third - bottom). So for \(\frac{b}{c}=\frac{e}{f}\)? Wait, but the problem's first proportion is \(\frac{b}{c}=\frac{\_}{f}\). Wait, maybe the first transversal (left) has segments \(b\) and \(c\), and the second transversal (right) has segments \(e\) and \(f\)? No, maybe I messed up. Wait, let's do the third proportion: \(\frac{c}{a + b}=\frac{f}{d + e}\)? Wait, no, the third proportion is \(\frac{c}{a + b}=\frac{f}{\_}\). \(a + b\) is the length from top to third parallel on the left (sum of \(a\) and \(b\)), so the corresponding length on the right is sum of \(d\) and \(e\) (sum of \(d\) and \(e\)). Wait, but the first two proportions:

First proportion: \(\frac{b}{c}=\frac{e}{f}\)? Wait, no, let's check the second proportion: \(\frac{d}{f}=\frac{a}{c}\)? Wait, no, maybe the correct correspondences are:

For \(\frac{b}{c}=\frac{e}{f}\)? Wait, no, the problem's first proportion: \(\frac{b}{c}=\frac{\_}{f}\). Let's look at the transversals. The left transversal: segments \(b\) (middle) and \(c\) (bottom part). The right transversal: segment \(e\) (middle) and \(f\) (bottom part). So \(\frac{b}{c}=\frac{e}{f}\)? Wait, but maybe the labels are different. Wait, maybe the left transversal has segments \(a\) (top), \(b\) (middle top), \(c\) (middle bottom), and the right has \(d\) (top), \(e\) (middle top), \(f\) (middle bottom). So by proportionality, \(\frac{b}{c}=\frac{e}{f}\), \(\frac{d}{f}=\frac{a}{c}\)? No, wait the second proportion: \(\frac{d}{f}=\frac{\_}{c}\). \(d\) is top part of right transversal, \(f\) is bottom part. On left, top part is \(a\), bottom part is \(c\). So \(\frac{d}{f}=\frac{a}{c}\)? Wait, no, maybe I got the segments wrong. Wait, let's start over.

The figure has four parallel lines (horizontal) cut by two transversals (slanted). The left transversal is divided into three segments: \(a\) (between first and second parallel), \(b\) (second and third), \(c\) (third and fourth). The right transversal is divided into three segments: \(d\) (first and second), \(e\) (second and third), \(f\) (third and fourth). By the theorem of parallel lines cutting transversals proportionally, \(\frac{a}{d}=\frac{b}{e}=\frac{c}{f}\), and also \(\frac{b}{c}=\frac{e}{f}\), \(\frac{d}{f}=\frac{a}{c}\)? Wait, no, \(\frac{d}{f}=\frac{a + b}{b + c}\)? No, the third proportion: \(\frac{c}{a + b}=\frac{f}{d + e}\). Wait, \(a + b\) is the length from first to third parallel on the left (sum of \(a\) and \(b\)), so the corresponding length on the right is sum of \(d\) and \(e\) (sum of \(d\) and \(e\)). So \(\frac{c}{a + b}=\frac{f}{d + e}\).

Wait, let's solve each proportion:

1. \(\frac{b}{c}=\frac{e}{f}\) (since \(b\) and \(e\) are middle segments, \(c\) and \(f\) are bottom segments)
2. \(\frac{d}{f}=\frac{a}{c}\) (since \(d\) and \(a\) are top segments, \(f\) and \(c\) are bottom segments)
3. \(\frac{c}{a + b}=\frac{f}{d + e}\) (since \(a + b\) is first to third on left, \(d + e\) is first to third on right)

Wait, but the problem's first proportion is \(\frac{b}{c}=\frac{\_}{f}\). So the missing term is \(e\). Second proportion: \(\frac{d}{f}=\frac{\_}{c}\), missing term is \(a\)? No, wait \(d\) is first to second on right, \(f\) is third to fourth. On left, first to second is \(a\), third to fourth is \(c\). So \(\frac{d}{f}=\frac{a}{c}\), so missing term is \(a\)? Wait, no, maybe I mixed up. Wait, let's check the third proportion: \(\frac{c}{a + b}=\frac{f}{d + e}\). So the denominator on the right is \(d + e\).

Wait, maybe the correct answers are:

First proportion: \(\frac{b}{c}=\frac{e}{f}\) (so missing is \(e\))

Second proportion: \(\frac{d}{f}=\frac{a}{c}\) (missing is \(a\))

Third proportion: \(\frac{c}{a + b}=\frac{f}{d + e}\) (missing is \(d + e\))

But let's verify with proportionality. If \(\frac{b}{c}=\frac{e}{f}\), cross - multiply: \(b\times f = c\times e\), which is true for parallel lines. For \(\frac{d}{f}=\frac{a}{c}\), cross - multiply: \(d\times c = f\times a\), which is true as \(\frac{a}{d}=\frac{c}{f}\) (from parallel lines). For the third, \(\frac{c}{a + b}=\frac{f}{d + e}\), cross - multiply: \(c(d + e)=f(a + b)\), and since \(\frac{a}{d}=\frac{b}{e}=\frac{c}{f}=k\) (constant), \(a = kd\), \(b = ke\), \(c = kf\). Substitute: \(kf(d + e)=f(kd + ke)\), which simplifies to \(kf(d + e)=kf(d + e)\), so it's true.

Step2: Fill in the Blanks

1. For \(\frac{b}{c}=\frac{\_}{f}\), the missing term is \(e\).
2. For \(\frac{d}{f}=\frac{\_}{c}\), the missing term is \(a\).
3. For \(\frac{c}{a + b}=\frac{f}{\_}\), the missing term is \(d + e\).

Answer:

1. \(\frac{b}{c}=\frac{\boldsymbol{e}}{f}\)
2. \(\frac{d}{f}=\frac{\boldsymbol{a}}{c}\)
3. \(\frac{c}{a + b}=\frac{f}{\boldsymbol{d + e}}\)