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Geometric Proof of Heron's Formula


Date: 01/25/2000 at 17:43:50
From: Joey Lloyd
Subject: How do I prove geometrically Hero(n)'s formula?

How do I prove Hero(n)'s formula using a circle with center P and 
radius R inscribed in triangle ABC?

Thank you so much.


Date: 01/26/2000 at 12:00:45
From: Doctor Floor
Subject: Re: How do I prove geometrically Hero(n)'s formula?

Hi, Joey,

Thanks for your question.

I know a method to prove Heron's formula geometrically with the help 
of the incircle of the triangle, but in this method I also use one of 
the excircles. I learned this method from Paul Yiu of Florida 
Atlantic University.

First, I have made a picture for you of the triangles and the two 
circles. Note that I have changed names: the incircle has center I and 
radius r (while the excircle (opposite to A) has center I' and radius 
r') instead of your center P and radius R.

  

In the following, the sides of the triangle are written as a = BC,
b = AC, and c = AB. I use s = (a+b+c)/2 for the semiperimeter.

1. From the fact that two tangents to a circle are congruent, we see 
   that AE = AG, CG = CF and BE = BF. So, for instance:

     AE+EB+CG = c+CG = s

   and thus CG = s-c.

   In the same way:

     CF = s-c
     AG = AE = s-a
     BE = BF = s-b

2. Again from the fact that two tangents to a circle are congruent we 
   see that AE' = AG, BE' = BJ and CG' = CJ. This gives us:

     AG'+AE' = AB+BJ+CJ+AC = 2s

   And we can conclude that AG' = AE' = s. And, for instance, 
   BE' = s-c.

3. Both I and I' lie on the internal angle bisector of <A.

   I' lies also on the external angle bisectors of <B and <C. These 
   external angle bisectors are perpendicular to the respective 
   internal angle bisectors. So BI and BI' are perpendicular; CI and 
   CI' are perpendicular as well.

4. From step 3 we can conclude that triangles EBI and E'I'B are 
   similar. From this we can conclude from E'I'/E'B = EB/EI:

      r'   s-b
     --- = ---
     s-c    r

   and thus 

     r*r' = (s-b)(s-c)   ....................[1]

5. We can also see that triangles AIG and AI'G' are similar. Here we 
   can conclude from IG/I'G' = AG/A'G' that:

     r    s-a
     -- = ---   .............................[2]
     r'    s

6. Multiplying [1] and [2] gives:

           (s-a)(s-b)(s-c)
     r^2 = ---------------
                  s

   and

              (s-a)(s-b)(s-c)
     r = sqrt(---------------)   ............[3]
                     s

7. It is not difficult to see that the area of ABC, let's call it K, 
   equals s*r. Combining this with [3] we find Heron's formula:

     K = sqrt(s(s-a)(s-b)(s-c))

I hope this helped. If you need more help, just write us back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Triangles and Other Polygons

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