Proof of Hero's formula

Date: 09/08/97 at 13:25:23
From: Russ Boyd
Subject: Hero's formula

Could you tell me where to find a proof of Hero's formula or help on 
how to derive it?  I am a teacher and my class is working on it.

Date: 09/08/97 at 16:05:14
From: Doctor Wilkinson
Subject: Re: Hero's formula

The book "Journey through Genius" has a proof very much like Hero's.  
You can derive it by brute force from the law of cosines.

-Doctor Wilkinson,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 09/08/97 at 16:35:37
From: Doctor Anthony
Subject: Re: Hero's formula

We use the formula  area = (1/2)bc.sin(A)

And sin(A) = 2 sin(A/2).cos(A/2)    so area = bc.sin(A/2).cos(A/2)

Next we find expressions for cos(A/2) and sin(A/2) in terms of 
s, a, b, c, where a, b, c are the sides of the triangle and s is 
the semi-perimeter.

So  s = (a+b+c)/2

                                    b^2 + c^2 - a^2
  cos(A) =   2cos^2(A/2) - 1    =   ----------------
                                         2bc


                                    b^2 + c^2 - a^2
                2cos^2(A/2) = 1 + ------------------
                                         2bc


                               (b+c)^2 - a^2      (b+c-a)(b+c+a)
                            =  --------------   = --------------
                                    2bc                2bc

                                 (2s-2a)2s
                            =   -----------
                                    2bc

                                   s(s-a) 
                     cos^2(A/2) = ---------
                                     bc

                                   b^2 + c^2 - a^2
Next  cos(A) = 1 - 2 sin^2(A/2) = ----------------
                                        2bc

                                       b^2 + c^2 - a^2
                    2sin^2(A/2) = 1 - ----------------
                                            2bc

                                   a^2 - (b-c)^2      (a-b+c)(a+b-c)
                                =  --------------  =  --------------
                                       2bc                  2bc

                                   (2s-2b)(2s-2c)
                                =  --------------
                                        2bc

                                    (s-b)(s-c)
                    sin^2(A/2)  =  ------------
                                         bc

 We can now return to the formula for the area of the triangle:

     (1/2)bc.sin(A) = bc.sin(A/2).cos(A/2) 

                      bc.sqrt[s(s-a)(s-b(s-c)]
                    = ------------------------
                               bc

                    = sqrt[s(s-a)(s-b)(s-c)]

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/