Question

consider the diagram and the proof below.
given: in △abc, ad ⊥ bc
prove: \\(\frac{\sin(b)}{b} = \frac{\sin(c)}{c}\\)

diagram of triangle abc with ad perpendicular to bc, labels: ac = b, ab = c, bc = a, ad = h, d between c and b

statement	reason
2. in △adb, \\(\sin(b) = \frac{h}{c}\\)	2. def of sine
3. \\(c\sin(b) = h\\)	3. mult. prop. of equality
4. in △acd, \\(\sin(c) = \frac{h}{b}\\)	4. def of sine
5. \\(b\sin(c) = h\\)	5. mult. prop. of equality
6.?	6. substitution
7. \\(\frac{\sin(b)}{b} = \frac{\sin(c)}{c}\\)	7. div. prop. of equality

what is the missing statement in step 6?

• \\(b = c\\)
• \\(\frac{h}{b} = \frac{h}{c}\\)
• \\(c\sin(b) = b\sin(c)\\)
• \\(b\sin(b) = c\sin(c)\\)

Explanation:

Step1: Analyze the given information

From step 2, we have \(\sin(B)=\frac{h}{c}\), so \(h = c\sin(B)\). From step 4, we have \(\sin(C)=\frac{h}{b}\), so \(h = b\sin(C)\).

Step2: Use substitution property

Since \(h = c\sin(B)\) and \(h = b\sin(C)\), by substitution (if \(a = b\) and \(a = c\), then \(b = c\) in the context of equality of \(h\) - related expressions), we get \(c\sin(B)=b\sin(C)\).

Answer:

\(c\sin(B) = b\sin(C)\) (the third option)