Geometric Proof of Heron's FormulaDate: 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/ |
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