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 $\frac{\sin(b)}{b} = \frac{\sin(c)}{c}$.
✓ 3. for $\triangle acd$, $\sin(c) = h/b$
✓ 4. multiply both sides of the equation by $b$ and simplify. the equation $\sin(c) = \frac{h}{b}$ becomes $b \sin(c)=h$

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

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

Explanation:

Step1: Recall prior derived equations

From steps for $\triangle ABD$, we get $c\sin(B) = h$; from step4, $b\sin(C) = h$.

Step2: Identify equality property

Since both equal $h$, use transitive property: if $a=h$ and $b=h$, then $a=b$.

Answer:

Transitive