How can you use the Cosine Rule to prove Heron's Formula?

We stat with the law of cosines, which we can assume to be true:

* c2 = a2 + b2 - 2ab*cos(C) Then rearrange it:

* cos(C) = a2 + b2 - c2/2ab Use the identity sin(x)=SQRT(1-cos(x))

* sin(C) = SQRT( 1 - (a2 + b2 - c2/2ab)2) Use the operator A = 1/2ab*sin(C) where A is area. Also, set one equal to 4a2b2 and factor it out.

* 2A/ab = SQRT(4a2b2 - (a2 + b2 - c2)2)/2ab ab's cancel, and the term inside the square root is the difference of two squares.

* A = 1/4*SQRT((2ab - (a2 + b2 - c2))(2ab + (a2 + b2 - c2))) when the two groups are simplified, the can be factored in binomial squares.

* A = 1/4*SQRT((c2 - (a - b)2)((a + b)2 - c2) Once again, we have differences of squares.

* A = 1/4*SQRT((c - (a - b))(c + (a - b))((a + b) - c)((a + b + c)) Simplify.

* A = 1/4*SQRT((c + b - a)(c + a - b)(a + b - c)(a + b +c)) Here comes the tricky part. We have four parts here. Three have two terms positive and one negative. Having a + b + c is like having the P. If we have a + b - c, that is like saying P - 2c, right? So to make it even easier, we can call s, the semi-perimeter, P/2. Then we can say a + b - c is 2s - 2c, or 2(s - c). We can apply that to all parts except the last one, which is just 2s.

* A = 1/4*SQRT(2(s - a)*2(s - b)*2(s - c)*2s) The two's can multiply together to 16 and come out of the root, canceling with the 1/4. we are left with good old Heron's formula.

* A = SQRT(s(s - a)(s - b)(s - c))