Answer:
Given: In ΔABC , [tex]AD \perp BC[/tex]
To prove that: [tex]\frac{\sin B}{b} =\frac{\sin C}{c}[/tex]
[tex]AD \perp BC[/tex] [Given]
In ΔADB
The sine angle is defined in the context of a right triangle is the ratio of the length of the side that is opposite that angle to the length of the longest side of the triangle.
[tex]\sin B =\frac{h}{c}[/tex] [By definition of sine] .....[1]
Multiplication Property of equality states that you multiply both sides of an equation by the same number.
Multiply by c to both sides of an equation [1] we get;
[tex]c \cdot \sin B =c \cdot\frac{h}{c}[/tex]
Simplify:
[tex]c \sin B = h[/tex] ......[2]
Now, In ΔACD
Using definition of sine:
[tex]\sin C =\frac{h}{b}[/tex]
Multiply both sides of an equation by b;
[tex]b \cdot \sin C = b \cdot \frac{h}{b}[/tex] [Multiplication Property of equality]
Simplify:
[tex]b \sin C = h[/tex] ......[3]
Substitute [3] in [2];
[tex]c \sin B = b \sin C[/tex] ......[4]
Division property of equality states that if you divide both sides of an equation by the same nonzero number the sides remains equal.
[4] ⇒[tex]\frac{\sin B}{b} =\frac{\sin C}{c}[/tex]
Therefore, the missing statement in step 6 is; [tex]c \sin B = b \sin C[/tex]
