Consider the diagram and the proof below. Given: In △ABC, AD ⊥ BC Prove: StartFraction sine (uppercase B) Over b EndFraction = StartFraction sine (uppercase C) Over c EndFraction Triangle A B C is shown. A perpendicular bisector is drawn from point A to point D on side C B forming a right angle. The length of A D is h, the length of C B is a, the length of C A is b, and the length of A B is c. A 2-column table has 7 rows. The first column is labeled Statement with entries In triangle A B C line segment A D is perpendicular to line segment B C, In triangle A D B sine (uppercase B) = StartFraction h Over c EndFraction, c sine (uppercase B) = h, In triangle A C D, sine (uppercase C) = StartFraction h Over b EndFraction, b sine (uppercase C) = h, question mark, StartFraction sine (uppercase B) Over b EndFraction = StartFraction sine (uppercase C) Over c EndFraction. The second column is labeled Reason with entries given, definition of sine, multiplication property of equality, definition of sine, multiplication property of equality, substitution, and division property of equality. What is the missing statement in Step 6? b = c StartFraction h Over b EndFraction = StartFraction h Over c EndFraction csin(B) = bsin(C) bsin(B) = csin(C)