Question

in the diagram of right triangle bcd below, \\(\overline{bc} \parallel \overline{fe}\\). which of the following ratios is equivalent to \\(\sin c\\)?

answer
\\(\bigcirc \frac{fe}{fd}\\) \\(\bigcirc \frac{fe}{ed}\\) \\(\bigcirc \frac{fd}{fe}\\) \\(\bigcirc \frac{fd}{ed}\\)

Explanation:

Step1: Recall Sine Definition

In right triangle \(BCD\), \(\sin C=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{BD}{CD}\) (wait, no, \( \angle B = 90^\circ \), so for \( \angle C \), opposite side is \(BD\), adjacent is \(BC\), hypotenuse \(CD\). But since \(BC \parallel FE\), triangles \(DFE\) and \(DBC\) are similar (AA similarity, as \( \angle D \) is common and \( \angle DFE = \angle DBC = 90^\circ \)).

Step2: Analyze Similar Triangles

In similar triangles \(DFE \sim DBC\), corresponding angles are equal. So \( \angle C = \angle DEF \). Now, in right triangle \(DFE\), \(\sin \angle DEF=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{FD}{ED}\)? Wait, no: in \( \triangle DFE \), right-angled at \(F\), \( \sin \angle DEF=\frac{FD}{ED}\)? Wait, no, \( \angle at E\): wait, \(FE \perp DB\), so \( \triangle DFE\) is right-angled at \(F\). So for \( \angle DEF \), opposite side is \(FD\), hypotenuse is \(ED\)? No, wait: \( \angle DEF \) has opposite side \(FD\), adjacent \(FE\), hypotenuse \(ED\). Wait, but \( \angle C = \angle DEF \) (since \(BC \parallel FE\), corresponding angles). So \( \sin C = \sin \angle DEF \).

Wait, let's re-express: In \( \triangle BCD \), right-angled at \(B\), \( \sin C = \frac{BD}{CD} \). In \( \triangle DFE \), right-angled at \(F\), \( \sin \angle DEF = \frac{FD}{ED} \)? No, wait, \( \angle DEF \) is equal to \( \angle C \), so \( \sin C = \sin \angle DEF \). Wait, no, maybe I mixed up. Wait, \(BC \parallel FE\), so \( \angle C = \angle DEF\) (alternate interior angles? Wait, \(BC \parallel FE\), transversal \(CD\), so \( \angle C = \angle DEF\)). Then in \( \triangle DFE \), right-angled at \(F\), \( \sin \angle DEF = \frac{FD}{ED}\)? No, \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\). For \( \angle DEF \), opposite side is \(FD\) (since \( \angle F = 90^\circ \), so \(FD\) is opposite \( \angle DEF\)), hypotenuse is \(ED\). Wait, but in \( \triangle BCD \), \( \sin C = \frac{BD}{CD} \). But since \( \triangle DFE \sim \triangle DBC\), \( \frac{FD}{BD}=\frac{FE}{BC}=\frac{ED}{CD}\). So \( \frac{FD}{ED}=\frac{BD}{CD}=\sin C \)? Wait, no: \( \sin C = \frac{BD}{CD} \), and \( \frac{FD}{ED}=\frac{BD}{CD}\) (from similarity: \( \frac{FD}{BD}=\frac{ED}{CD} \implies \frac{FD}{ED}=\frac{BD}{CD} \)). Wait, no, cross-multiplying \( \frac{FD}{BD}=\frac{ED}{CD} \implies FD \cdot CD = BD \cdot ED \implies \frac{BD}{CD}=\frac{FD}{ED}\). So \( \sin C = \frac{BD}{CD} = \frac{FD}{ED} \)? Wait, no, let's check the options. The options are \( \frac{FE}{FD} \), \( \frac{FE}{ED} \), \( \frac{FD}{FE} \), \( \frac{FD}{ED} \). Wait, maybe I made a mistake. Let's re-express \( \sin C \) in \( \triangle BCD \): \( \sin C = \frac{BD}{CD} \). In \( \triangle DFE \), \( \sin \angle DEF = \frac{FD}{ED} \), and \( \angle DEF = \angle C \), so \( \sin C = \frac{FD}{ED} \)? Wait, no, \( \angle DEF \): in \( \triangle DFE \), right-angled at \(F\), \( \angle DEF \) has opposite side \(FD\), hypotenuse \(ED\), so \( \sin \angle DEF = \frac{FD}{ED} \). Since \( \angle C = \angle DEF \), then \( \sin C = \frac{FD}{ED} \)? Wait, but let's check the other option: \( \frac{FE}{ED} \)? No, \(FE\) is adjacent. Wait, maybe I messed up the angle. Wait, \(BC \parallel FE\), so \( \angle C = \angle FED\) (alternate interior angles). In \( \triangle DFE \), right-angled at \(F\), \( \sin \angle FED = \frac{FD}{ED} \)? Wait, no, \( \angle FED \): the sides: \(FE\) is horizontal, \(FD\) is vertical, \(ED\) is hypotenuse. So \( \sin \angle FED = \frac{FD}{ED} \), yes. So \( \sin C = \sin \angle FED = \frac{FD}{ED} \)? Wait, but let's check the options. The options are:

1. \( \frac{FE}{FD} \)
2. \( \frac{FE}{ED} \)
3. \( \frac{FD}{FE} \)
4. \( \frac{FD}{ED} \)

Wait, maybe I made a mistake in the angle. Wait, \( \angle C \) in \( \triangle BCD \): \( \sin C = \frac{BD}{CD} \). In \( \triangle DFE \), \(BD\) corresponds to \(FD\), \(CD\) corresponds to \(ED\), so \( \frac{BD}{CD} = \frac{FD}{ED} \), so \( \sin C = \frac{FD}{ED} \). Wait, but let's check the other way: in \( \triangle DFE \), \( \sin \angle D = \frac{FE}{ED} \), but \( \angle D \) is common. Wait, no, \( \angle C + \angle D = 90^\circ \), and \( \angle DEF + \angle D = 90^\circ \), so \( \angle C = \angle DEF \), so \( \sin C = \cos \angle D \), but maybe that's not helpful. Wait, let's re-express \( \sin C \): in \( \triangle BCD \), \( \sin C = \frac{BD}{CD} \). In \( \triangle DFE \), \( \frac{FD}{ED} = \frac{BD}{CD} \) (by similarity, since \( \triangle DFE \sim \triangle DBC \), so corresponding sides are proportional: \( \frac{FD}{BD} = \frac{FE}{BC} = \frac{ED}{CD} \), so cross-multiplying \( \frac{FD}{ED} = \frac{BD}{CD} \)). Therefore, \( \sin C = \frac{FD}{ED} \), which is one of the options (the last one: \( \frac{FD}{ED} \)). Wait, but let's check the options again. The options are:

- \( \frac{FE}{FD} \)
- \( \frac{FE}{ED} \)
- \( \frac{FD}{FE} \)
- \( \frac{FD}{ED} \)

So the correct ratio should be \( \frac{FD}{ED} \)? Wait, no, wait: in \( \triangle DFE \), right-angled at \(F\), \( \sin \angle DEF = \frac{FD}{ED} \), and \( \angle DEF = \angle C \), so \( \sin C = \frac{FD}{ED} \). So the answer is \( \frac{FD}{ED} \).

Wait, but let's check again. Maybe I mixed up opposite and adjacent. Let's define \( \angle C \): in \( \triangle BCD \), right-angled at \(B\), so \( \angle C \) has opposite side \(BD\), adjacent \(BC\), hypotenuse \(CD\). So \( \sin C = \frac{BD}{CD} \). In \( \triangle DFE \), right-angled at \(F\), \( \angle DEF = \angle C \), so \( \sin \angle DEF = \frac{FD}{ED} \) (since \(FD\) is opposite \( \angle DEF\), \(ED\) is hypotenuse). Since \( \triangle DFE \sim \triangle DBC \), \( \frac{BD}{CD} = \frac{FD}{ED} \), so \( \sin C = \frac{FD}{ED} \). So the correct option is \( \frac{FD}{ED} \).

Answer:

\(\frac{FD}{ED}\) (the last option: \( \boldsymbol{\frac{FD}{ED}} \))