Question

two right triangles, $\triangle abd$ and $\triangle acd$, were created from $\triangle abc$ by constructing $\overline{ad} \perp \overline{cb}$.
follow these steps to prove that $\frac{\sin(b)}{b} = \frac{\sin(c)}{c}$.
$\sin(c) = \frac{h}{b}$ becomes

1. by the property, we can set the expressions equal to each other.

$c \sin(b) = b \sin(c)$

1. divide both sides of the equation by and simplify to get

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

Explanation:

Step1: Identify target equation

We start with $c \sin(B) = b \sin(C)$

Step2: Divide by $bc$

Divide both sides by $bc$ to isolate the ratios:
$\frac{c \sin(B)}{bc} = \frac{b \sin(C)}{bc}$

Step3: Simplify the fractions

Cancel common terms in numerator and denominator:
$\frac{\sin(B)}{b} = \frac{\sin(C)}{c}$

Answer:

The value to fill in the blank is $bc$ (or $b \cdot c$). The final proven identity is $\frac{\sin(B)}{b} = \frac{\sin(C)}{c}$