270
QUADRILATÈRE
![{\displaystyle \left.{\begin{array}{rll}x^{2}&={\cfrac {bc\left(a^{2}+d^{2}\right)+ad\left(b^{2}+c^{2}\right)}{bc+ad}}&={\cfrac {(ac+bd)(ab+cd)}{bc+ad}},\\\\y^{2}&={\cfrac {ab\left(c^{2}+d^{2}\right)+cd\left(a^{2}+b^{2}\right)}{ab+cd}}&={\cfrac {(ac+bd)(bc+ad)}{ab+cd}}\,;\end{array}}\right\}\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/00783d4b21f3bfe48936dc48d3f84072ef47a08e)
(1)
et par suite
![{\displaystyle xy=ac+bd,\qquad {\frac {x}{y}}={\frac {ab+cd}{bc+ad}}.\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1c878449b5e6ad3f690b24d45d64c009286b38c)
(2)
Si, dans l’expression
![{\displaystyle \operatorname {Cos} .(a,d)={\frac {a^{2}+d^{2}-x^{2}}{2ad}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f645c47a3589b1a41de464fb1e646d1f0da99ad)
on met pour
la valeur que nous venons d’obtenir, il viendra, toutes réductions faites,
![{\displaystyle \operatorname {Cos} .(a,d)={\frac {a^{2}+d^{2}-b^{2}-c^{2}}{2(bc+ad)}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50e9f924671f1c16e61d255fe5b5d9c4e8cf295a)
or, on a
![{\displaystyle 2\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)=1-\operatorname {Cos} .(a,d),\quad 2\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)=1+\operatorname {Cos} .(a,d)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41bb753f0b199bb56a884e4465361f30e233dec7)
donc, en substituant et divisant par ![{\displaystyle 2,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/993d53082bc05c266933af9a892e1ce2128547cd)
![{\displaystyle {\begin{array}{rll}\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)&={\cfrac {(b+c)^{2}-(a-d)^{2}}{4(bc+ad)}}&={\cfrac {(b+c+d-a)(a+b+c-d)}{4(bc+ad)}},\\\\\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)&={\cfrac {(a+d)^{2}-(b-c)^{2}}{4(bc+ad)}}&={\cfrac {(a+c+d-b)(d+a+b-c)}{4(bc+ad)}},\,;\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4d888ce57f9cfdf4d6b008aa822de889fd253b1)
mais on a d’ailleurs
![{\displaystyle \operatorname {Sin} .(a,d)=2\operatorname {Sin} .{\frac {1}{2}}(a,d)\operatorname {Cos} .{\frac {1}{2}}(a,d)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10c207e303542e965371714576f17d3989a635d0)
substituant donc et posant, pour abréger,