Consider the diagram and the proof below.
Given: In AABC, AD 1 BC
Prove:
sin(B)
sino
b
Reason
given
del of sine
2
Statement
In SABC, AD 18C
2
In SADB, sin(B) -
3 csin(B) -
4
IN SACD, sin(C)-
5. sin(C)-
2
Sin(B) sin(C)
3
4
mult prop of quality
del of sine
A
8
5
8
7
mult prop. of equality
substitution
di prop of equality
b
c
What is the missing statement in Step 6?
h
b=0
a
D
В
csin(B) = bsin(C)
bsin(B) = csin(C)

Step-by-step explanation:

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