Heron and Bretschneider

The Heron relation gives area Δ of triangle from its sides a, b, c.
Naming s the semiperimeter s = (a+b+c)/2
 Δ = √( s(s-a)(s-b)(s-c) ) 

This relation is generalized for a quadrilateral into Bretschneider relation :

 Δ = √( (s-a)(s-b)(s-c)(s-d) - abcd.cos²((B+D)/2) ) 

d = 0 gives back Heron relation in triangle, it is then sufficient to prove Bretschneider relation, although there is a direct elementar proof of Heron relation.

A further consequence : For a quadrilateral with given sides a, b, c, d, the maximal area is when this quadrilateral is inscribed (B+D = 180°), term in cos² is then null, and the relation becomes Brahmagupta relation.

Proof of Bretschneider relation :
The law of cosinus in triangle ABC gives AC² = a² + b² - 2ab cosB, and similarily in ACD AC² = c² + d² - 2cd cosD hence : a² + b² - 2ab cosB = c² + d² - 2cd cosD or : (a² + b² - c² - d²)² = (2ab cosB - 2cd cosD)² = 4a²b²cos²B + 4c²d²cos²D - 8abcd cosB cosD [1]

Also ABCD area is Δ = Area(ABC) + Area(ACD) = 1/2 ab sinB + 1/2 cd sinD that is 16Δ² = (2ab sinB + 2cd sinD)² = 4a²b²sin²B + 4c²d²sin²D + 8abcd sinB sinD [2]

Adding [1] and [2] together, and because sin² + cos² = 1 and cosB cosD - sinBsinD = cos(B + D) : (a² + b² - c² - d²)² +16Δ² = 4a²b² + 4c²d² - 8abcd cos(B+D)

Completing the square 4a²b² + 4c²d² : (a² + b² - c² - d²)² +16Δ² = (2ab + 2cd)² - 8abcd(1 + cos(B+D)) = (2ab + 2cd)² - 16abcd cos²((B+D)/2) as 1 + cos(t) = 2.cos²(t/2)
That is 16Δ² = ((2ab + 2cd)² - (a² + b² - c² - d²)²) - 16abcd cos²((B+D)/2)

The first term can also be written (2ab + 2cd + a² + b² - c² - d²)(2ab + 2cd - a² - b² + c² + d²) that is ((a + b)² - (c - d)²)((c + d)² - (a - b)²) or
(a + b + c - d)(a + b - c + d)(c + d + a - b)(c + d - a + b)
and naming s = (a + b + c + d)/2 : a + b + c - d = 2(s - d) etc
And the product is 16(s - a)(s - b)(s - c)(s - d)
hence Δ² = (s - a)(s - b)(s - c)(s - d) - abcd cos²((B+D)/2)
QED

Direct proof of Heron relation

Consider triangle ABC, with sides AB = c, AC = b and BC = a, with altitude AH = h.
Area of this triangle is Δ = BC.AH / 2 = a.c.sin(B) / 2 hence sin(B) = 2Δ/(a.c)
Also the relation b² = a² + c² - 2 a.c.cos(B) gives cos(B) = (a² + c² - b²)/(2 a.c)
cos²(B) + sin²(B) = 1 is then (a² + c² - b²)² + 16Δ² = 4 a²c² and factoring :
16Δ² = (2a.c + a² + c² - b²)(2a.c - a² - c² + b²) = ((a+c)² - b²)(b² - (a-c)²) and finally :
16Δ² = (a+b+c)(a-b+c)(a+b-c)(b+c-a) QED.

Another proof

Consider the incircle touching the sides in D, E, F, and the excircle touching sides in K,M,N.
The equal tangents BF and BD (resp AF=AE, CD=CE) gives (AB+BC+AC) = 2(BF+AE+CE) = 2(BF+AC) hence  BF = BD = p-b  and similarily  AF = AE = p-a et CD = CE = p-c 
AM = AN = (AM+AN)/2 = ((AB+BK) + (AC+CK))/2 = (a+b+c)/2 = p
hence  BM = BK = p-c, CN = CK = p-b 

Area of BIC = a.r /2 and similarily (AIC) = b.r /2 and (AIB) = c.r /2
Area of ABC is then  Δ = (a+b+c).r /2 = p.r 

Also with the excircle :
Area (ABC) = (AJM) + (AJN) - (BJM) - (BJK) - (CJK) - (CJN) = 2(AJM) - 2(JBC)
 Δ = (p-a).ra 

Hence Δ² = p(p-a).r.ra

Consider right triangles BIF and JBM, they are similar (BI_|_BJ) and then IF/BM = BF/JM that is r.ra = IF.JM = BM.BF = (p-c)(p-b)

Finally results Δ² = p(p-a)(p-b)(p-c) QED.

Note : The previous relations give at once :

r² = (p-a)(p-b)(p-c)/p and
ra² = p(p-b)(p-c)/(p-a)

 

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