Question

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

1. in △abc, ad⊥bc.
2. in △adb, $sin(b)=\frac{h}{c}$.
3. $csin(b)=h$
4. in △acd, $sin(c)=\frac{h}{b}$.
5. $bsin(c)=h$

6.?

1. $\frac{sin(b)}{b}=\frac{sin(c)}{c}$

reason

1. given
2. def. of sine
3. mult. prop. of equality
4. def. of sine
5. mult. prop. of equality
6. substitution
7. div. prop. of equality

what is the missing statement in step 6?
$b = c$
$\frac{h}{b}=\frac{h}{c}$
$csin(b)=bsin(c)$
$bsin(b)=csin(c)$

Explanation:

Step1: Review prior steps

From Step 3: \( c\sin(B) = h \); Step 5: \( b\sin(C) = h \)

Step2: Substitute h

Since both equal h, set equal: \( c\sin(B) = b\sin(C) \)

Answer:

C. \( c\sin(B) = b\sin(C) \)