Notation
Let a = |AB|, b = |BC|, c = |CD|, d = |DA|, p = |AC|, and q = |BD|.
Let R and s be the circumradius, and semiperimeter of ABCD
respectively. Let [PQ...UV] denote the area of polygon PQ...UV.
Solution
s = (a+b+c+d)/2 (1)
For each vertex of ABCD, construct two segments from the vertex
to the boundary of the convex hull such that the segments are
perpendicular to the two sides of ABCD adjacent to the vertex.
It is easy to see that the area of the convex hull consists of
[ABCD] surrounded by four circular sectors and four rectangles
with the same width R. The four circular sectors form a circular
disk of radius R. The area of the covex hull is
[ABCD] + PI*R2 + R*(2s) (2)
All we need now is [ABCD] and R in terms of a, b, c, and d.
Finding [ABCD]
p2 = a2 + b2 - 2*a*b*cos(∠ABC)
= c2 + d2 - 2*c*d*cos(∠ADC)
= c2 + d2 + 2*c*d*cos(∠ABC)
(a2 + b2) - (c2 + d2) = 2*(a*b + c*d)*cos(∠ABC)
[(a2 + b2) - (c2 + d2)]2/16 =
[(a*b + c*d)*cos(∠ABC)]2/4 (3)
[ABCD] = [ABC] + [ADC
= a*b*sin(∠ABC)/2 + c*d*sin(∠ADC)
= (a*b + c*d)*sin(∠ABC)/2
[ABCD]2 = [(a*b + c*d)*sin(∠ABC)]2/4 (4)
[ABCD]2 + [(a2 + b2) - (c2 + d2)]2/16
= (a*b + c*d)2/4 (3)+(4)
[ABCD]2 = (a*b + c*d)2/4 -
[(a2 + b2) - (c2 + d2)]2/16
We could stop here, but with a lot more algebra we get
Brahmagupta's formula:
[ABCD] = √[ (s-a)(s-b)(s-c)(s-d) ] (5)
Finding R
Lemma: If ΔXYZ has circumradius R and side lengths
x, y, and z, then
[XYZ] = x*y*z/(4*R).
Proof: Let M be the midpoint of XY and O the circumcenter
of ΔXYZ.
[XYZ] = x*y*sin(∠XZY)/2
= x*y*sin(∠XOY/2)/2
= x*y*sin(∠XOM)/2
= x*y*( [z/2]/R )/2
= x*y*z/(4*R)
The circumradius of ABCD is the same as the circumradius of
ΔABC, ΔADC, ΔBAD, and ΔBCD:
[ABCD] = [ABC] + [ADC]
= a*b*p/(4*R) + c*d*p/(4*R)
= p*(a*b + c*d)/(4*R)
[ABCD] = [BAD] + [BCD]
= a*d*q/(4*R) + b*c*q/(4*R)
= q*(a*d + b*c)/(4*R)
[ABCD]2 = p*q*(a*b + c*d)*(a*d + b*c)/(16*R2)
From Ptolemy's theorem, p*q = a*c + b*c
[ABCD]2 = (a*b + c*d)*(a*c + b*c)*
(a*d + b*c)/(16*R2)
R2 = (a*b + c*d)*(a*c + b*c)*
(a*d + b*c)/(16*[ABCD]2)
R = √[ (a*b + c*d)*(a*c + b*c)*(a*d + b*c) ]/
(4*[ABCD]) (6)
Equations (1), (2), (5), and (6) satisfy the problem.
QED
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