Question:

Step 1: Express terms using the semi-perimeter. Let $s = \frac{a+b+c}{2}$ be the semi-perimeter of the triangle. Then the terms in the numerator can be expressed as: $a+b+c = 2s$ $b+c-a = (a+b+c) - 2a = 2s - 2a = 2(s-a)$ $c+a-b = (a+b+c) - 2b = 2s - 2b = 2(s-b)$ $a+b-c = (a+b+c) - 2c = 2s - 2c = 2(s-c)$ Step 2: Substitute these into the given expression. $$ \frac{(2s) \cdot 2(s-a) \cdot 2(s-b) \cdot 2(s-c)}{4b^2c^2} = \frac{16s(s-a)(s-b)(s-c)}{4b^2c^2} = \frac{4s(s-a)(s-b)(s-c)}{b^2c^2} $$ Step 3: Apply Heron's formula for the area of a triangle. The area of a triangle, denoted by $\Delta$, is given by Heron's formula: $$ \Delta = \sqrt{s(s-a)(s-b)(s-c)} $$ Therefore, $s(s-a)(s-b)(s-c) = \Delta^2$. Step 4: Substitute $\Delta^2$ into the expression. The expression becomes: $$ \frac{4\Delta^2}{b^2c^2} $$ Step 5: Express the area of a triangle using the sine rule. The area of a triangle can also be expressed as $\Delta = \frac{1}{2}bc\sin A$. Squaring this, we get: $$ \Delta^2 = \left(\frac{1}{2}bc\sin A\right)^2 = \frac{1}{4}b^2c^2\sin^2 A $$ Step 6: Substitute this expression for $\Delta^2$ back into the simplified expression. $$ \frac{4 \cdot \left(\frac{1}{4}b^2c^2\sin^2 A\right)}{b^2c^2} = \frac{b^2c^2\sin^2 A}{b^2c^2} = \sin^2 A $$