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| | ==The Proof== | | ==The Proof== |
| − | Denote the area of the quadrilateral by ''S''. Then we have
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| − | :<math> \begin{align} S &= \text{area of } \triangle ADB + \text{area of } \triangle BDC \\
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| − | &= \tfrac{1}{2}pq\sin A + \tfrac{1}{2}rs\sin C
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| − | \end{align} </math>
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| − | Therefore
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| − | :<math> 4S^2 = (pq)^2\sin^2 A + (rs)^2\sin^2 C + 2pqrs\sin A\sin C. \, </math>
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| − | The [[law of cosines]] implies that
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| − | :<math> p^2 + q^2 -2pq\cos A = r^2 + s^2 -2rs\cos C, \, </math>
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| − | because both sides equal the square of the length of the diagonal ''BD''. This can be rewritten as
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| − | :<math>\tfrac14 (r^2 + s^2 - p^2 - q^2)^2 = (pq)^2\cos^2 A +(rs)^2\cos^2 C -2 pqrs\cos A\cos C. \,</math>
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| − | Substituting this in the above formula for <math>4S^2</math> yields
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| − | :<math>4S^2 + \tfrac14 (r^2 + s^2 - p^2 - q^2)^2 = (pq)^2 + (rs)^2 - 2pqrs\cos (A+C). \, </math>
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| − | This can be written as
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| − | :<math>16S^2 = (r+s+p-q)(r+s+q-p)(r+p+q-s)(s+p+q-r) - 16pqrs \cos^2 \frac{A+C}2. </math>
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| − | Introducing the semiperimeter
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| − | :<math>T = \frac{p+q+r+s}{2},</math>
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| − | the above becomes
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| − | :<math>16S^2 = 16(T-p)(T-q)(T-r)(T-s) - 16pqrs \cos^2 \frac{A+C}2</math>
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| − | and Bretschneider's formula follows.
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| − | NOTE TO ALL: this proof was taken from Wikipedia on December the 1st, 2006.
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| | ==See Also== | | ==See Also== |
| | * [[Brahmagupta's formula]] | | * [[Brahmagupta's formula]] |
| | * [[Geometry]] | | * [[Geometry]] |
Revision as of 09:03, 2 December 2006
(in that order) and 
. Bretschneider's formula states that the ![$[ABCD]=\frac{1}{4}*\sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}$](http://latex.artofproblemsolving.com/7/5/9/7596249e635c730943d0f5493ab8e0d9f2399293.png)
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