Question

in △abc, ef || ab. what is the ratio of bc to fc?

Explanation:

Step1: Identify Similar Triangles

Since \( EF \parallel AB \), by the Basic Proportionality Theorem (Thales' theorem), \( \triangle CEF \sim \triangle CAB \). So, the ratios of corresponding sides are equal.

Step2: Find Length of \( CE \) and \( CA \)

From the diagram, \( CE = 4 \) and \( EA = 3 \), so \( CA = CE + EA = 4 + 3 = 7 \).

Step3: Apply Proportionality

For similar triangles \( \triangle CEF \) and \( \triangle CAB \), \( \frac{CE}{CA}=\frac{CF}{CB} \). Let \( CF = x \) and \( CB = y \), then \( \frac{4}{7}=\frac{x}{y} \), but we need \( \frac{BC}{FC}=\frac{y}{x} \). Taking reciprocal of \( \frac{4}{7}=\frac{x}{y} \), we get \( \frac{y}{x}=\frac{7}{4} \)? Wait, no, wait. Wait, \( CE = 4 \), \( EA = 3 \), so \( CA = 7 \). Wait, actually, the ratio of \( CB \) to \( FC \) should be based on the ratio of \( CA \) to \( CE \) because \( EF \parallel AB \), so \( \frac{CB}{FC}=\frac{CA}{CE} \). Since \( CA = CE + EA = 4 + 3 = 7 \), \( CE = 4 \), so \( \frac{CB}{FC}=\frac{7}{4} \)? Wait, no, maybe I mixed up. Wait, the problem is ratio of \( BC \) to \( FC \). Let's re - examine.

Wait, \( EF \parallel AB \), so \( \triangle CEF \sim \triangle CAB \) (by AA similarity, since \( \angle C \) is common and \( \angle CEF=\angle CAB \) as \( EF \parallel AB \), corresponding angles). So the ratio of sides \( \frac{CE}{CA}=\frac{CF}{CB} \). We know \( CE = 4 \), \( EA = 3 \), so \( CA=CE + EA = 4 + 3=7 \). So \( \frac{CE}{CA}=\frac{4}{7}=\frac{CF}{CB} \), then \( \frac{CB}{CF}=\frac{7}{4} \)? But the options seem to have \( \frac{7}{4} \)? Wait, maybe the options are mis - written or I misread. Wait, maybe the length of \( CE \) is 4 and \( EA \) is 3, so \( CA = 7 \), and since \( EF\parallel AB \), \( \frac{CF}{CB}=\frac{CE}{CA}=\frac{4}{7} \), so \( \frac{CB}{CF}=\frac{7}{4} \).

Wait, maybe the options are: Let's check the options again. The options are \( \frac{7}{4} \)? Wait, maybe the original problem has \( CE = 4 \), \( EA = 3 \), so \( CA=7 \), and \( \frac{BC}{FC}=\frac{CA}{CE}=\frac{7}{4} \).

Answer:

\(\frac{7}{4}\) (assuming the options include this, if the options are like \( \frac{7}{4} \) among them)