Date: Mon, 15 Dec 1997 17:20:58 -0500 (EST) From: John Conway <conway@math.Princeton.

EDU> Subject: Heron To: Peter Doyle <doyle@mathgrad.ucsd.edu> I thought about proofs of Heron's formula and came up with a few, but couldn't find your email address so didn't send them to you. To my mind the "real" proof says (somehow) that the squared area is a quadratic in a^2, b^2, c^2 that vanishes when +-a+-b+-c = 0, and so ... . I found several ways of proving the first half of the sentence, but non of them were really pleasing, and you will doubtless prefer your own. The other way I like is to prove the half-angle formulae, (they are very easy from the geometry), and then use area = bc sin(A/2)cos(A/2).

I also looked in my "Workman" and found, as I had vaguely remembered, two distinct analogues of Heron for spherical triangles. The neater one was l'Huiller's, but the other was also interesting. I'll try to bring Workman into Fine Hall, so that next time I can quote it exactly. [I like his occasional remarks to the student about what should be learned. I'll quote some of those too!] JHC

From doyle Wed Dec 17 19:47:46 1997 To: conway@math.princeton.edu, res@math.umd.edu Subject: Heron's formula Content-Length: 1471 John, How about this: If a quadratic form takes values A, B, C for the three vectors of a superbase, then the determinant of the form relative to the superbase is 1/4 (2 A B + 2 A C + 2 B C - A^2 - B^2 - C^2) (Symmetric of degree 2; vanishes when A=0, B=C.) Any triangle congruent to the image of the standard equilateral triangle with vertices {1, omega, omega^2} under a linear mapping. Its squared area is proportional to the determinant of the quadratic form Q you get by pulling back the standard norm in the image under this linear mapping. This pulled-back form Q takes values A=a^2, B=b^2, C=c^2 for the three vectors of the superbase {I, I omega, I omega^2}. Thus the determinant of Q is 1/4 (2 A B + 2 A C + 2 B C - A^2 - B^2 - C^2) = 1/4 (b+c-a) (a+b-c) (a+c-b) (a+b+c)

and the squared area of the triangle is 1/16 (b+c-a) (a+b-c) (a+c-b) (a+b+c) This can be seen directly, without computation, by noting that the determinant of a quadratic form is certainly some quadratic function of the norms of any three independent vectors, and hence the area of a triangle is quadratic in a^2, b^2, c^2, and vanishes when +-a+-b+-c=0. Brahmagupta's formula says that the squared area of a quadrilateral inscribed in a circle is 1/16 (a+b+c-d)(a+b+d-c)(a+c+d-b)(b+c+d-a) This reduces to Heron's formula when d=0. To verify this formula, we need only show that the squared area is a quartic function of the side lengths. Why is this true, I wonder? Peter Date: Tue, 23 Dec 1997 15:02:44 -0500 (EST) From: John Conway <conway@math.Princeton.EDU> Subject: Re: Heron's formula To: Peter Doyle <doyle@euclid.ucsd.edu> Cc: res@math.umd.edu On Wed, 17 Dec 1997, Peter Doyle wrote: > John, > > How about this: > > If a quadratic form takes values A, B, C for the three vectors of a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> > Thus the determinant of Q is > > 1/4 (2 A B + 2 A C + 2 B C - A^2 - B^2 - C^2) > = > 1/4 (b+c-a) (a+b-c) (a+c-b) (a+b+c) > > and the squared area of the triangle is > > 1/16 (b+c-a) (a+b-c) (a+c-b) (a+b+c) > > This can be seen directly, without computation, by noting that the > determinant of a quadratic form is certainly some quadratic function > of the norms of any three independent vectors, and hence the area of a > triangle is quadratic in a^2, b^2, c^2, and vanishes when +-a+-b+-c=0. Well, ... yes, I suppose so. I found a few other reasons to justify the quadraticity-in-squares, at least one of which seemed easier. > Brahmagupta's formula says that the squared area of a quadrilateral inscribed > in a circle is > > 1/16 (a+b+c-d)(a+b+d-c)(a+c+d-b)(b+c+d-a)

> > This reduces to Heron's formula when d=0. You may not know the generalization to arbitrary quadrilaterals, namely area^2 = (s-a)(s-b)(s-c)(s-d) - ?abcd.cos^2(u)

where u is the semi-sum of two opposite angles. (This shows, inter alia, that the area is maximized in the cyclic case.) > To verify this formula, we > need only show that the squared area is a quartic function of the > side lengths. Why is this true, I wonder? > > Peter Dunno at the moment, but surely this can't be hard. I have a nice little book that gives some other nice stuff about cyclic quadrilaterals. Namely, by putting a,b,c,d into the three essentially different orders, we obtain 3 such, whose diagonals have only 3 lengths, which my book calls e,f,g (in the wrong order, but I'll use the right one). Then e^2 = (bd+ca)(cd+ab)/(ad+bc) etc., and I think that area is something like efg/4R, from which (when corrected) you can deduce a formula for R. [This is only a sample.] Ptolemy's theorem shows that AD.BC, BD.CA, CD.AB are the sides of a triangle, which degenerates to a straight line precisely when the quadrilateral ABCD is cyclic. I would love to know just how this triangle is related to ABCD - for instance do its angles have any geometrical interpretation. Which quadrilaterals correspond to the same triangle? I think I've seen a TINY bit of this stuff, but would like to know it ALL. JHC Date: Mon, 10 May 1999 09:42:17 -0400 (EDT) From: John Conway <conway@math.Princeton.EDU> To: Peter Doyle <doyle@euclid.ucsd.edu> Subject: triangle trigonometry Peter - you once asked did I know a nice proof of Heron's formula, to which the answer was that I didn't. Well, I still don't, but (in connection with my proposed "Triangle Book") am feeling a bit happier, because I've worked out a really nice ab initio development of trigonometry. By "the Somecenter picture" I mean the picture obtained by joining the Somecenter to the vertices, and dropping perps from it to the sides. Then in the incenter picture the sides get cut into pieces of lengths sa,sb,sc say (sx is really s_x), giving the equations

sb + sc = a sc + sa = b sa + sb = c which we semi-add to get sa + sb + sc = (a+b+c)/2 = s, whence say,

sa = s-a, sb = s-b, sc = s-c.

In the circumcenter picture, the sides get bisected into pieces of lengths RsinA, RsinB, RsinC, so we get the sine rule a/sinA = b/sinB = c/sinC = 2R. In the orthocenter picture, the b-edge is cut into pieces of lengths a.cosC and c.cosA, so we get ( 2Rsin(C+A) = ) 2RsinB = 2R( sinA.cosC + sinC.cosA ) whence the addition-formula for sines, and we can similarly get the addition-formula for cosines by looking at the way the altitude splits up. But also in that picture draw squares on the sides, and continue the altitudes through them, so cutting the b-square into pieces of areas bc.cosA and ab.cosC = SA and SB, say. (SX is really S_X). Then we get the equations SB + SC = aa SC + SA = bb SA + SB = cc, which semi-add to give SA + SB + SC + (aa+bb+cc)/2 = S, say, whence SA = S - aa = (bb + cc - aa)/2, the cosine formula.

Now we go into algebraic mode. ssa = (b+c)^2/4 - aa/4 sbsc = aa/4 - (b-c)^2/4 = =

We find

(bc + bc.cosA)/2 = bc.cos^2(A/2) (bc - bc.cosA)/2 = bc.sin^2(A/2)

whence the half-angle formulae sin(A/2) = root(sbsc/bc) cos(A/2) = root(ssa/bc) tan(A/2) = root(sbsc/ssa) and also, taking the geometric mean, Heron's formula: root(ssasbsc) = bc.sin(A/2).cos(A/2) = bc.sinA/2 = DELTA A pretty neat treatment, eh? But I'd still like to reduce the algebraic part, and would absolutely LOVE to get a "symmetrical" proof of Heron's formula. JHC

Date: Fri, 3 Mar 2000 12:03:49 -0500 (EST) From: John Conway <conway@math.Princeton.EDU> To: "Peter G. Doyle" <doyle@emmy.Dartmouth.EDU> Subject: Heron

You were once keen to know the best proof of Heron. I've found is this: DELTA^2 = ro.so.ra.sa = so.sa.sb.sc

The best

since from the diagram below we have ra/sb = sc/ro, the triangles Ia X C and C Y Io being similar in view of the angles Io C V = C/2 and Ia C X the complement of this. Ia | | |ra Io | |ro ----X----C-------Y------A sb sc I write so,sa,sb,sc for the usual s,s-a,s-b,s-c, and remark that the formulae DELTA = rx.sx (x=0,a,b,c) have well-known very simple proofs. It would be nice to find a simple symmetric 4-dimensional proof, but I've not yet managed to do so. Regards, JHC

From doyle Wed Jun 28 10:17:04 2000 To: conway Subject: Cantor's worm-eaten paradise John, I liked your Heron proof very much. Brahmagupta's formula? Can you modify it to get

I've just got a new digital tape recorder, and I've been copying the Conway tapes. Looking them over, I can't believe I didn't copy them earlier. I mean, what if my house had burned down? Of course I want to make the tapes available on the web. But the first concern is to assure that the priceless wisdom on these tapes is preserved. I came across the following snippet, delivered after the memorable `ordinal walk' field trip in your `Romance of Numbers' class. PGD: Tell me about the benefit of being brought up in a religious household. JHC: Well, in a religious household of this particular kind of religion, anyway. Well, Cantor's construction is so fantastic---so beautiful---so incredible, that we ought to know what it is. If we start disbelieving it before we've even heard it through, you know, we haven't given it a chance, so to speak. I don't believe it, I really don't believe it... But I think it's very sound policy to lie. I mean, let's remember

Groucho Marx's wonderful thing, `If you can fake sincerity you've got it made.' Let's fake sincerity for a bit. And then, you know, after a time we can tell them it's all a bit dubious. PGD: It's a Cantorian paradise, right?

JHC: Yes: `Out of this paradise that Cantor has created nobody will expell us.' PGD: JHC: PGD: Hilbert said that? Yeah. Oh, those were the days...

JHC: Well, nobody actually has thrown us out! All they've just done---I mean, it's like the problem of America, basically. I mean, there are drug addicts and shooting in the streets. It's dangerous, this paradise. It's got contradictions all over the place, OK? It just a slightly sort of worm-eaten paradise. ... Come on, switch that thing off! At the time now I think you believe astonishing Peter From conway@math.Princeton.EDU Wed Jun 28 14:28:09 2000 Date: Wed, 28 Jun 2000 14:27:09 -0400 (EDT) From: John Conway <conway@math.Princeton.EDU> To: "Peter G. Doyle" <doyle@hilbert.dartmouth.edu> Subject: Re: Cantor's worm-eaten paradise I was too young to appreciate what you were saying, but I understand. Like horseshoes, Cantor's theory works whether in it or not. And the less you believe in it, the more it is how well it works.

On Wed, 28 Jun 2000, Peter G. Doyle wrote: > I liked your Heron proof very much. > Brahmagupta's formula? Can you modify it to get

It's not really my proof - I found it in Casey's "Sequel to Euclid" (of around 1880) and he says upfront that he allows himself to quote stuff from the following list of books without attribution, so it's probably in several of them, and can be regarded as traditional. No, I don't see how to get Brahmagupta (but am still hoping). > PGD: Tell me about the benefit of being brought up in a religious household. ... > It just a slightly sort of worm-eaten paradise. > that thing off! Why didn't you?! > At the time I was too young to appreciate what you were saying, but > now I think I understand. Like horseshoes, Cantor's theory works whether ... Come on, switch

> you believe in it or not. And the less you believe in it, the more > astonishing it is how well it works. I like the "like horseshoes"! JHC From doyle Wed Jun 28 15:06:51 2000 To: conway Subject: Horseshoes John, I did turn the machine off, just as you instructed, and thereby deprived posterity of any further pearls you had to cast. The phrase `like horseshoes' is a reference to the story about the quantum mechanic who had a horseshoe nailed above his door. Someone said, `Surely, Prof. Bohr (?), you don't believe in horseshoes?' `No,' said Bohr, `but I understand they work whether you believe in them or not.' Peter From conway@math.Princeton.EDU Wed Jun 28 17:38:10 2000 Date: Wed, 28 Jun 2000 17:37:14 -0400 (EDT) From: John Conway <conway@math.Princeton.EDU> To: "Peter G. Doyle" <doyle@hilbert.dartmouth.edu> Subject: Re: Horseshoes On Wed, 28 Jun 2000, Peter G. Doyle wrote: > > > > > > > > > > John, I did turn the machine off, just as you instructed, and thereby deprived posterity of any further pearls you had to cast. The phrase `like horseshoes' is a reference to the story about the quantum mechanic who had a horseshoe nailed above his door. Someone said, `Surely, Prof. Bohr (?), you don't believe in horseshoes?' `No,' said Bohr, `but I understand they work whether you believe in them or not.' Yes, I think you've told me this before. Peter - I read "our" paper on division by 3 for the first time when I was at Northwestern earlier this year, and was disturbed by the way that the point got buried in the fluff. I'll expand on this if you like (but must then naturally allow myself to speak as sharply and sarcastically as I should if we were together!), but basically just want to say that we should get together some time (soon, please!) to write a short and snappy version to be published on permanent paper. I hope you like this plan. How on earth can we hope to prove Brahmagupta? One plan is to show somehow that the answer has to be the square root of a polynomial of a certain form, then remark that its difference from s(s-a)(s-b)(s-c) must vanish when [so-and-so], so must be a multiple of [such-and-such], and then the multiplier must be 1. I imagine I could force through such a proof, and maybe it's the best we can expect?

JHC PS: I'm writing a book on the geometry of the triangle (which, by the way, will be the second triangular book that I'll own!), and I'm sure you'll like some of the goodies in there. That's another thing to discuss when we're next together. J From doyle Wed Jun 28 20:05:46 2000 To: conway Subject: Fluff John, Yes, I agree that there is a lot of fluff in the `preliminary version' of the division by 3 paper. I've been meaning to take it out for a long time now. The draft version is on my web page under the heading `in remission'. It was originally listed under the heading `in preparation', but at some point I decided that this was an exaggeration. This draft has just recently gotten some press in John Baez's column `This Week's Finds in Mathematical Physics': http://math.ucr.edu/home/baez/week147.html Two things that deserve to be thought about: (1) In the draft I claim that similar methods will show that if 3a=2b then there is some c so that a=2c and b=3c. However, the last time I thought about this I didn't see how to do it, and I'm not quite sure whether we really ever did figure this out. (2) I believe that this method of dividing by three does not require the power-set axiom. I would like to know if this is indeed the case. Are you free to come visit in Hanover, say for a week or so? We could make a whole bunch more TAPES! Or I could come visit in Princeton, and we could make a whole bunch of TAPES. Peter From conway@math.Princeton.EDU Wed Jun 28 22:23:12 2000 Date: Wed, 28 Jun 2000 22:21:52 -0400 (EDT) From: John Conway <conway@math.Princeton.EDU> To: "Peter G. Doyle" <doyle@hilbert.dartmouth.edu> Subject: Re: Fluff On Wed, 28 Jun 2000, Peter G. Doyle wrote: > Yes, I agree that there is a lot of fluff ... > ... I've been meaning to take it out for a long time now. Dear Mr Fluff I regret to inform you that it isn't possible for you to do that alone. > > > > > > Two things that deserve to be thought about: (1) In the draft I claim that similar methods will show that if 3a=2b then there is some c so that a=2c and b=3c. However, the last time I thought about this I didn't see how to do it, and I'm not quite sure whether we really ever did figure this out.

I think we did, and if not, that we can and should. > (2) I believe that this method of dividing by three does not require > the power-set axiom. I would like to know if this is indeed the case. Certainly feels that way - but it's the sort of thing that's best checked through in tandem rather than alone. > Are you free to come visit in Hanover, say for a week or so? We could > make a whole bunch more TAPES! Or I could come visit in Princeton, > and we could make a whole bunch of TAPES. Dear Mr Tapeworm, Thank you very much for your kind invitation. social secretary when I'll be free to accept. Your most humble and obedient master. From conway@Math.Princeton.EDU Wed Jan 17 12:38:27 2001 Date: Wed, 17 Jan 2001 12:39:13 -0500 (EST) From: John Conway <conway@Math.Princeton.EDU> To: "Peter G. Doyle" <doyle@hilbert.dartmouth.edu> Subject: Re: Visit? On Thu, 11 Jan 2001, Peter G. Doyle wrote: > John, > > I understand from Wendy and Bob that you might be induced to visit Dartmouth > for a week or so sometime at the end of January. Dartmouth has money to > pay expenses (train, hotel, meals). Or are you all booked up now? > If not now, when, and how do I contact your social secretary? I'd love to come, but am afraid I can't do so at the end of January. I hope we can fix up something a bit later. You remember asking for the right proof of Heron's formula? Well, here's a pretty nice one: usual way that First you prove in the obvious and I shall ask my

ro.so = ra.sa = rb.sb = rc.sc where so = (a+b+c)/2, sa = so - a, etc., are the radii of the in- and ex- circles. Then you prove (eg.) sb.sc = ro.ra and ro, ra, rb, rc

by similar triangles Ic

B Io --------------C-------X---------A-----Y------------

Namely, in triangle

Io X A

we have A Y Ic

Io X = ro, we have

X A = sa

while in the similar triangle

A Y = sb, Y Ic = rc.

Then DELTA^2 = (ro.so)(ra.sa) = (so.sa)(ro.ra) = (so.sa)(sb.sc). I agree this hasn't got the symmetry we'd like to see, and nor does it do something elegant in four dimensions, as we'd hoped, but I think you'll agree it's pretty good. [It seems to have been the standard proof in the long-ago days when Heron's theorem had a proof.] John Conway

Heron's Formula
An important theorem in plane geometry, also known as Hero's formula. Given the lengths of the sides , , and and the semiperimeter (1) of a triangle, Heron's formula gives the area of the triangle as (2) Heron's formula may be stated beautifully using a Cayley-Menger determinant as

(3)

Another highly symmetrical form is given by

(4)

(Buchholz 1992).

Expressing the side lengths , , and in terms of the radii , , and ' of the mutually tangent circles centered on the triangle vertices (which define the Soddy circles), (5) (6) (7) gives the particularly pretty form (8) Heron's proof (Dunham 1990) is ingenious but extremely convoluted, bringing together a sequence of apparently unrelated geometric identities and relying on the properties of cyclic quadrilaterals and right triangles. Heron's proof can be found in Proposition 1.8 of his work Metrica (ca. 100 BC-100 AD). This manuscript had been lost for centuries until a fragment was discovered in 1894 and a complete copy in 1896 (Dunham 1990, p. 118). More recently, writings of the Arab scholar Abu'l Raihan Muhammed al-Biruni have credited the formula to Heron's predecessor Archimedes prior to 212 BC (van der Waerden 1961, pp. 228 and 277; Coxeter and Greitzer 1967, p. 59; Kline 1990; Bell 1986, p. 58; Dunham 1990, p. 127). A much more accessible algebraic proof proceeds from the law of cosines, (9) Then (10) giving (11) (12) (13) (14) (15) (Coxeter 1969). Heron's formula contains the Pythagorean theorem as a degenerate case.

Formulas for Finding Area of A Triangle

SAS Formula Heron's Formula

Area of Triangle Worksheets Area of Triangle Calculator

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Heron's Formula

Most schoolchildren know the traditional formula used to find the area of a triangle: A = (1/ 2)*b*h. Obviously, to find the area of a triangle using this formula, one must know the length of a side of the triangle (the base, b) and the length of the altitude to that side (the height, h). On the other hand, Heron's Formula can be used to find the area of a triangle when one knows the lengths of the three sides. Note that it is not necessary to know a height in order to use this formula. Heron's Formula states: Given the lengths a, b, and c of the three sides of a triangle...

...and after finding the semiperimeter, s, of the triangle,...

...the area of the triangle can be found using this formula:

Click here for a Geometer's SketchPad file to manipulate and to relate the traditional area formula to Heron's Formula.

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Proofs of Heron's Formula

There are many proofs of Heron's Formula. Most can be categorized as algebraic, geometric, or trigonometric. The following list includes a presentation of one proof of each of these types. Click here to see an algebraic proof of Heron's Formula.

Click here to see a geometric proof of Heron's Formula. This proof is based on Heron's proof of the formula in Metrica. Click here to see a trigonometric proof of Heron's Formula.

An Algebraic Proof of Heron's Formula by Shannon Umberger

Note: This proof was adapted from a proof on Kevin Brown's MathPages website. http://www.seanet.com/~ksbrown/kmath196.htm

As in the above figure, given a triangle, let the three sides have lengths a, b, and c. Construct the altitude to side c, and let that altitude have length h. Let the distance from the foot of the altiude to the other endpoint of side a be d, and let the distance from the foot of the altitude to the other endpoint of side b be e. By segment addition, it can be stated that c = d + e. Using the resulting right triangles and the Pythagorean Theorem, it can be stated that d2 + h2 = a2 and e2 + h2 = b2. By the subtraction property of equality, d2 + h2 - (e2 + h2) = a2 - b2. It follows that d2 - e2 = a2 - b2. Using the division property of equality, since c = d + e, then (d2 - e2)/ (d + e) = (a2 - b2)/ c, which implies that d - e = (a2 - b2)/ c. Using the addition property of equality and c = d + e gives d - e + (d + e)= [(a2 - b2)/ c] + c. Then 2d = (a2 - b2 + c2)/ c. Solving for d gives d = (a2 b2 + c2)/ 2c. Using the traditional formula to find the area of a triangle, the given triangle has area A = hc/ 2. Since d2 + h2 = a2, then h = (a2 - d2)1/2. After substituting into the area formula, A = (1/

2)*c*(a2 - d2)1/2 = (1/ 2)*(a2c2 - d2c2)1/2 = (1/ 2)*[(ac)2 - (dc)2]1/2. Plugging in the above value for d gives A = (1/ 2)*[(ac)2 - (((a2 - b2 + c2)/ 2c)*c)2]1/2 = (1/ 2)*[(ac)2 - ((a2 - b2 + c2)/ 2)2]1/2. The above equation is equivalent to the equation A = (1/ 2)*[(4(ac)2 - (a2 - b2 + c2)2)/ 4]1/2. Factoring out 1/ 4 gives A = (1/ 2)*(1/ 2)*[(2ac)2 - (a2 - b2 + c2)2]1/2. So it can be stated that 4A = [(2ac)2 - (a2 - b2 + c2)2]1/2; squaring both sides gives 16A2 = (2ac)2 - (a2 - b2 + c2)2. Factoring the right side of the equation gives 16A2 = [2ac + (a2 - b2 + c2)]*[2ac - (a2 - b2 + c2)]. After some regrouping, 16A2 = [(a2 + 2ac + c2) - b2]*[b2 - (a2 - 2ac + c2)]. Factoring again, 16A2 = [(a + c)2 - b2]*[b2 - (a - c)2]. Finally, factoring once more gives 16A2 = [(a + c) + b]*[(a + c) - b]*[b + (a - c)]*[b - (a - c)] = (a + b + c)(a - b + c)(a + b - c)(-a + b + c). Since the semiperimeter of the above triangle is defined as s = (a + b + c)/ 2, it follows that 2s = a + b + c, and so 2s - a = b + c. Adding -a to both sides of the equation gives 2s - 2a = -a + b + c. Therefore, 2(s - a) = -a + b + c. Similar algebra steps give 2(s - b) = (a - b + c) and 2(s c) = (a + b - c). If follows by substitution that 16A2 = 2s*[2(s - b)]*[2(s - c)]*[2(s - a)]. A little rearranging gives 16A2 = 16s(s - a)(s - b)(s - c), which implies A2 = s(s - a)(s - b)(s - c). Thus, A = [s(s a)(s - b)(s - c)]1/2, which was to be proved. QED.

A Geometric Proof of Heron's Formula by Shannon Umberger

Note: This proof was adapted from the outline of a proof on page 194 in the 6th edition of An Introduction to the History of Mathematics by Howard Eves.

Given triangle ABC, let the length of segment BC be a, the length of segment AC be b, and the length of segment AB be c. Note the perimeter, p, of triangle ABC = a + b + c. Half of the perimeter is called the semiperimeter, s, and so for triangle ABC, s = (a + b + c)/ 2.

Let the point of intersection of the angles bisectors of angles A, B, and C be called point I. Construct segments AI, BI, and CI.

Next, let the point of intersection of the perpendicular to side AC through point I be called point D, the point of intersection of the perpendicular line to side BC through point I be called point E, and the point of intersection of the perpendicular line to side AB through point I be called point F.

By definition, point I is called the incenter of triangle ABC. The circle with point I as the center and that passes through points D, E, and F is called the incircle of triangle ABC.

Now, since segment IB is the angle bisector of angle B, then angles IBF and IBE are equal. Since angles IFB and IEB are right by construction, then they are equal, and thus angles BIF and BIE are equal (since all angles of a triangle sum to 180 degrees). Note that segments IF and IE are equal since both are radii of the same circle. Therefore, triangles BIF and BIE (in red) are congruent by SAS.

Similarly, triangles CID and CIE (in green) are congruent by SAS.

Also similarly, triangles AIF and AID (in blue) are congruent by SAS.

Because corresponding sides of congruent triangles are congruent, it follows that segments BF and BE (in red) are equal, segments CD and CE (in green) are equal, and segments AD and AF (in blue) are equal.

By using the formula that states the area of a triangle equals one-half the base times the height, the area of triangle AIB = (1/ 2)*AB*IF,...

...the area of triangle BIC = (1/ 2)*BC*IE,...

...and the area of triangle AIC = (1/ 2)*AC*ID.

Therefore, the area of triangle ABC is the sum of the areas of triangles AIB, BIC, and AIC.

By substitution, the area of triangle ABC = (1/ 2)*AB*IF + (1/ 2)*BC*IE + (1/ 2)*AC*ID. Since segments IF, IE, and ID are equal (they are all radii of the same circle), it follows with a little algebra that the area of triangle ABC = (1/ 2)*ID*(AB + BC + AC). But AB + BC + AC = a + b + c, and so this sum is really the perimeter of triangle ABC; therefore, the area of triangle ABC = (1/ 2)*ID*p. Since (1/ 2)*p = s, the semiperimeter of triangle ABC, then the area of triangle ABC = ID*s.

Now, construct point G such that G lies on the same line as segment AC, and segment CG equals segment BE. Remember p = a + b + c = AB + BC + AC, and so by segment addition

and substitution, p = (BE + CE) + (CD + AD) + (AF + BF) = (BF + BE) + (CE + CD) + (AD + AF). Furthermore, by substitution, p = 2*BE + 2*CD + 2*AD = 2(BE + CD + AD). Again, by substitution, p = 2(CG + CD + AD). But, AG = AD + DC + CG by segment addition, and so by substitution, p = 2*AG. Therefore, s = AG.

Thus, by substitution, the area of triangle ABC = ID*AG.

Construct a line perpendicular to segment AI through point I (in pink) and a line perpendicular to segment AG through point C (in light blue). Let the point of intersection of these two lines be called point H. Let the point of intersection of segment IH and segment AG be called point J. Construct segment AH (in yellow).

Since angles AIH and ACH are right (by construction), then triangles AIH and ACH are right. Note the triangles share a common hypotenuse, segment AH. It follows that these triangles are inscribed in a common circle with segment AH as the diameter of the circle.

This fact means that quadrilateral AICH is cyclic, and so opposite angles AIC and AHC are supplementary.

Since the sum of the angles at point I is 360 degrees, by angle addition, (angle BIF + angle BIE) + (angle CIE + angle CID) + (angle AID + angle AIF) = 360 degrees. By substitution, 2*(angle BIE) + 2*(angle CID) + 2*(angle AID) = 360 degrees, and so angle BIE + angle CID + angle AID = 180 degrees. But angle CID + angle AID = angle AIC by angle addition, thus angle BIE + angle AIC = 180 degrees, and so angles BIE and AIC are supplementary.

Since angles BIE and AHC are supplementary to the same angle, it follows that angle BIE = angle AHC.

Because angles BIE and AHC are equal, and angles BEI and ACH are equal (both are right by construction), then triangles BIE and AHC are similar by AA similarity.

By substitution, AC/CG = AC/BE. Then, by the definition of similar triangles, AC/BE = HC/IE. Now, because angles IJD and HJC are vertical angles, they are equal. Also, angles IDJ and HCJ are equal (both are right by construction). Therefore, triangles IJD and HJC are similar by AA similarity.

By substitution, HC/IE = HC/ID. Then, by the definition of similar triangles, HC/ID = CJ/DJ. Thus, by transitivity, AC/CG = CJ/DJ.

The next few steps of the proof require a few algebra tricks. Since AC/CG = CJ/DJ, then AC/CG + 1 = CJ/DJ + 1.

So AC/CG + CG/CG = CJ/DJ + DJ/DJ. This implies that (AC + CG)/CG = (CJ + DJ)/DJ. But by segment addition, AC + CG = AG, and CJ + DJ = CD. Therefore, by substitution, AG/CG = CD/DJ. Well, then, AG/CG * 1 = CD/DJ * 1. So AG/CG * AG/AG = CD/DJ * AD/AD. Then AG2/CG*AG = CD*AD/DJ*AD.

Now, since triangle AIJ is right (by construction), and D is a point on the hypotenuse, then DJ*AD = ID2 (by the geometric mean).

So, by substitution, AG2/CG*AG = CD*AD/ID2. By cross multiplication, AG2*ID2 = CG*AG*CD*AD. It follows that AG*ID = [CG*AG*CD*AD]1/ 2. But, the area of triangle ABC = AG*ID. Thus, by transitivity, the area of triangle ABC = [CG*AG*CD*AD]1/ 2.

Finally, remember that segment AG = s.

Note that segment CG = AG - AC by segment addition, and so CG = s - b by substitution. Also note that segment CD = AG - (AD + CG) by segment addition. So CD = AG - (AF + BF) by substitution, and so CD = s - c by substitution. And note that segment AD = AG - (CD + CG) by segment addition. So AD = AG - (CE + BE) by substitution, and so AD = s - a by substitution.

Therefore, by substitution, the area of triangle ABC = [s*(s - a)*(s - b)*(s - c)]1/ 2. QED.

A Trigonometric Proof of Heron's Formula by Shannon Umberger

Note: This proof was adapted from a proof on Dr. Jim Wilson's website (Department of Mathematics Education, College of Education, University of Georgia). http://jwilson.coe.uga.edu/emt725/Heron/Trig.Heron.html

Given a triangle with sides of lengths a, b, and c. Trigonometry gives an area formula that states A = (1/ 2)absinC, where C is the angle opposite side c. Using the Law of Cosines, c2 = a2 + b2 - 2abcosC. Solving for cosC gives cosC = (a2 + b2 c2)/ 2ab. Rearrange the Pythagorean identity (sinC)2 + (cosC)2 = 1 to state sinC = [1 (cosC)2]1/2. By substitution, sinC = [1 - ((a2 + b2 - c2)/ 2ab)2]1/2. Factoring gives sinC = [(1 + (a2 + b2 - c2)/ 2ab)*(1 - (a2 + b2 - c2)/ 2ab)]1/2. It follows that sinC = [((2ab + a2 + b2 - c2)/ 2ab)*((2ab - a2 - b2 + c2)/ 2ab)]1/2. Factoring out the denominator gives sinC = (1/ 2ab)*[(2ab + a2 + b2 - c2)(2ab - a2 - b2 + c2)]1/2. After some regrouping, sinC = (1/ 2ab)*[((a2 + 2ab + b2) - c2)(c2 - (a2 - 2ab + b2))]1/2. Factoring gives sinC = (1/ 2ab)*[((a + b)2 - c2)(c2 - (a - b)2)]1/2. After factoring once more, sinC = (1/ 2ab)*[((a + b) + c)((a + b) - c)(c + (a - b))(c - (a - b))]1/2. With some rearranging, sinC = (1/ 2ab)*[(a + b + c)(a + b - c)(a - b + c)(-a + b + c)]1/2. Since the semiperimeter of the above triangle is defined as s = (a + b + c)/ 2, it follows that 2s = a + b + c, and so 2s - a = b + c. Adding -a to both sides of the equation gives 2s - 2a = -a + b + c. Therefore, 2(s - a) = -a + b + c. Similar algebra steps give 2(s - b) = (a - b + c) and 2(s c) = (a + b - c). If follows by substitution that sinC = (1/ 2ab)*[2s*[2(s - c)]*[2(s - b)]*[2(s - a)]]1/2. With some rearranging, sinC = (1/ 2ab)*[16s(s - a)(s - b)(s - c)]1/2 = (1/ 2ab)*4*[s(s - a)(s - b)(s c)]1/2 = (2/ ab)*[s(s - a)(s - b)(s - c)]1/2.

Now, substitute the value of sinC into the area formula A = (1/ 2)absinC. So, A = (1/ 2)*ab*(2/ ab)*[s(s - a)(s - b)(s - c)]1/2. Thus, A = [s(s - a)(s - b)(s - c)]1/2, which was to be proved. QED.

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Acknowledgement First and foremost, I would like to thank the principle of Sekolah Menengah KebangsaanPerempuan St. George, Pn. Shariffah Afifah for giving me the permission to do this AdditionalMathematics Project Work.Secondly, I would also like to show my gratitude towards the endless guidance andsupport given by my Additional Mathematics subject teacher, Mr. Ooi Chong Keat throughoutthe process of doing this project work. He taught me and my friends patiently on ways toaccomplish this project work.Apart from that, I would like to acknowledge and range my earnest appreciation to myparents for giving me supports in every ways, such as money, to buy anything that are related tothis project work and their advise, which is the most needed for this project.Lastly, a special thank you to all my friends for aiding me in solving some of thecalculations. They were helpful to share information and ideas in order to get this project work done.And especially to God, who made all things possible.Thank you, everyone.

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