Editing Heron's formula (section)

== Proofs ==
There are many ways to prove Heron's formula, for example using [[trigonometry]] as below, or the [[incenter]] and one [[excircle]] of the triangle,<ref>{{Cite web|date=15 December 1997|title=Personal email communication between mathematicians John Conway and Peter Doyle|url=https://math.dartmouth.edu/~doyle/docs/heron/heron.txt|access-date=25 September 2020}}</ref> or as a special case of [[De Gua's theorem]] (for the particular case of acute triangles),<ref>{{Cite journal|last=Lévy-Leblond|first=Jean-Marc|date=2020-09-14|title=A Symmetric 3D Proof of Heron's Formula|journal=The Mathematical Intelligencer|volume=43|issue=2|pages=37–39|language=en|doi=10.1007/s00283-020-09996-8|issn=0343-6993|doi-access=free}}</ref> or as a special case of [[Brahmagupta's formula]] (for the case of a degenerate cyclic quadrilateral).

===Trigonometric proof using the law of cosines===
A modern proof, which uses [[algebra]] and is quite different from the one provided by Heron, follows.<ref>{{cite book
| author=Niven, Ivan
| title=Maxima and Minima Without Calculus
| url=https://archive.org/details/maximaminimawith0000nive
| url-access=registration
| publisher=The Mathematical Association of America
| year=1981
| pages=[https://archive.org/details/maximaminimawith0000nive/page/7 7–8]}}</ref>
Let {{tmath|a,}} {{tmath|b,}} {{tmath|c}} be the sides of the triangle and {{tmath|\alpha,}} {{tmath|\beta,}} {{tmath|\gamma}} the [[angle]]s opposite those sides.
Applying the [[law of cosines]] we get

<math display=block>\cos \gamma = \frac{a^2+b^2-c^2}{2ab}</math>

[[File:Triangle with notations 2 without points.svg|thumb|A triangle with sides {{mvar|a}}, {{mvar|b}} and {{mvar|c}}]]
From this proof, we get the algebraic statement that

<math display=block>\sin \gamma = \sqrt{1-\cos^2 \gamma} = \frac{\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2 }}{2ab}.</math>

The [[altitude (triangle)|altitude]] of the triangle on base {{tmath|a}} has length {{tmath|b\sin\gamma}}, and it follows

<math display=block>\begin{align}
A
&= \tfrac12 (\mbox{base}) (\mbox{altitude}) \\[6mu]
&= \tfrac12 ab\sin \gamma \\[6mu]
&= \frac{ab}{4ab}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2} \\[6mu]
&= \tfrac14\sqrt{-a^4 - b^4 - c^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 b^2 c^2} \\[6mu]
&= \tfrac14\sqrt{(a + b + c)(- a + b + c)(a - b + c)(a + b - c)} \\[6mu]
&= \sqrt{
  \left(\frac{a + b + c}{2}\right)
  \left(\frac{- a + b + c}{2}\right)
  \left(\frac{a - b + c}{2}\right)
  \left(\frac{a + b - c}{2}\right)} \\[6mu]
&= \sqrt{s(s-a)(s-b)(s-c)}.
\end{align}</math>

===Algebraic proof using the Pythagorean theorem===
[[Image:Triangle with notations 3.svg|thumb|270px|Triangle with altitude {{mvar|h}} cutting base {{mvar|c}} into {{math|''d'' + (''c'' − ''d'')}}]]
The following proof is very similar to one given by Raifaizen.<ref>{{Cite journal
  | last = Raifaizen
  | first = Claude H.
  | title = A Simpler Proof of Heron's Formula
  | journal = Mathematics Magazine
  | volume = 44
  | number = 1
  | pages = 27–28
  | year = 1971
| doi = 10.1080/0025570X.1971.11976093
 }}</ref>
By the [[Pythagorean theorem]] we have <math>b^2 = h^2 + d^2</math> and <math>a^2 = h^2 + (c - d)^2</math> according to the figure at the right. Subtracting these yields <math>a^2 - b^2 = c^2 - 2cd.</math> This equation allows us to express {{tmath|d}} in terms of the sides of the triangle:
<math display=block>d = \frac{-a^2 + b^2 + c^2}{2c}.</math>
For the height of the triangle we have that <math>h^2 = b^2 - d^2.</math> By replacing {{tmath|d}} with the formula given above and applying the [[difference of squares]] identity we get
<math display=block>
\begin{align}
 h^2 &= b^2-\left(\frac{-a^2 + b^2 + c^2}{2c}\right)^2 \\
     &= \frac{(2bc - a^2 + b^2 + c^2)(2bc + a^2 - b^2 - c^2)}{4c^2} \\
     &= \frac{\big((b + c)^2 - a^2\big)\big(a^2 - (b - c)^2\big)}{4c^2} \\
     &= \frac{(b + c - a)(b + c + a)(a + b - c)(a - b + c)}{4c^2} \\
     &= \frac{2(s - a) \cdot 2s \cdot 2(s - c) \cdot 2(s - b)}{4c^2} \\
     &= \frac{4s(s - a)(s - b)(s -c )}{c^2}.
\end{align}
</math>

We now apply this result to the formula that calculates the area of a triangle from its height:
<math display=block>\begin{align}
 A &= \frac{ch}{2} \\
   &= \sqrt{\frac{c^2}{4} \cdot \frac{4s(s - a)(s - b)(s - c)}{c^2}} \\
   &= \sqrt{s(s - a)(s - b)(s - c)}.
\end{align}</math>

===Trigonometric proof using the law of cotangents===
[[File:Herontriangle2greek.svg|thumb|270px|right|Geometrical significance of {{math|''s'' − ''a''}}, {{math|''s'' − ''b''}}, and {{math|''s'' − ''c''}}.  See the [[law of cotangents]] for the reasoning behind this.]]
If {{tmath|r}} is the radius of the [[incircle]] of the triangle, then the triangle can be broken into three triangles of equal altitude {{tmath|r}} and bases {{tmath|a,}} {{tmath|b,}} and {{tmath|c.}} Their combined area is
<math display=block>A = \tfrac12ar + \tfrac12br + \tfrac12cr = rs,</math>
where <math>s = \tfrac12(a + b + c)</math> is the semiperimeter.

The triangle can alternately be broken into six triangles (in congruent pairs) of altitude {{tmath|r}} and bases {{tmath|s - a,}} {{tmath|s - b,}} and {{tmath|s - c}} of combined area (see [[law of cotangents]])
<math display=block>\begin{align}
 A &= r(s-a) + r(s-b) + r(s-c) \\[2mu]
   &= r^2\left(\frac{s - a}{r} + \frac{s - b}{r} + \frac{s - c}{r}\right) \\[2mu]
   &= r^2\left(\cot{\frac{\alpha}{2}} + \cot{\frac{\beta}{2}} + \cot{\frac{\gamma}{2}}\right) \\[3mu]
   &= r^2\left(\cot{\frac{\alpha}{2}} \cot{\frac{\beta}{2}} \cot{\frac{\gamma}{2}}\right)\\[3mu]
   &= r^2\left(\frac{s - a}{r} \cdot \frac{s - b}{r} \cdot \frac{s - c}{r}\right) \\[3mu]
   &= \frac{(s-a)(s-b)(s-c)}{r}.
\end{align}</math>

The middle step above is <math display=inline>\cot{\tfrac{\alpha}{2}} + \cot{\tfrac{\beta}{2}} + \cot{\tfrac{\gamma}{2}} = {}</math><math>\cot{\tfrac{\alpha}{2}}\cot{\tfrac{\beta}{2}}\cot{\tfrac{\gamma}{2}},</math> the [[Proofs of trigonometric identities#Miscellaneous – the triple cotangent identity|triple cotangent identity]], which applies because the sum of half-angles is <math display=inline>\tfrac\alpha2 + \tfrac\beta2 + \tfrac\gamma2 = \tfrac\pi2.</math>

Combining the two, we get
<math display=block>A^2 = s(s - a)(s - b)(s - c),</math>
from which the result follows.