Consider the diagram and the derivation below. Given: In △ABC, AD ⊥ BC Derive a formula for the area of △ABC using angle C. Triangle A B D is shown. A perpendicular bisector is drawn from point A to point D on side C B. The length of C B is a, the length of B A is c, the length of C A is b, and the length of A D is h. It is given that in △ABC, AD ⊥ BC. Using the definition of sine with angle C in △ACD results in sin(C) = StartFraction h Over b EndFraction. Using the multiplication property of equality to isolate h, the equation becomes bsin(C) = h. Knowing that the formula for the area of a triangle is A = One-halfbh is and using the side lengths as shown in the diagram, which expression represents the area of △ABC? One-halfbsin(C) absin(C) One-halfcbsin(C) One-halfhbsin(C)