277
INSCRIT.
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .^{2}{\frac {1}{2}}(a,d)=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b+c+d)\operatorname {Cos} .{\frac {1}{4}}(a+c-b-d)\operatorname {Sin} .{\frac {1}{4}}(a+b+c-d)\operatorname {Sin} .{\frac {1}{4}}(b+c+d-a)}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}},\\\\&\operatorname {Cos} .^{2}{\frac {1}{2}}(a,d)=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b-c-d)\operatorname {Cos} .{\frac {1}{4}}(b+c-a-d)\operatorname {Sin} .{\frac {1}{4}}(d+a+b-c)\operatorname {Sin} .{\frac {1}{4}}(c+d+a-b)}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}}.\\\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bdf9c26c2e6adf1975bcb5e40b1e678453f32b35)
En faisant, pour abréger,
![{\displaystyle {\begin{aligned}&\operatorname {Cos} .{\frac {1}{4}}(a+b+c+d)\operatorname {Cos} .{\frac {1}{4}}(a+b-c-d)\operatorname {Cos} .{\frac {1}{4}}(a+c-b-d)\operatorname {Cos} .{\frac {1}{4}}(b+c-a-d)=M\,;\\\\&\operatorname {Sin} .{\frac {1}{4}}(b+c+d-a)\operatorname {Sin} .{\frac {1}{4}}(c+d+a-b)\operatorname {Sin} .{\frac {1}{4}}(d+a+b-c)\operatorname {Sin} .{\frac {1}{4}}(a+b+c-d)=N\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52a96d3646cd5f4f6309b258ab4fe56b5f139d0b)
auquel cas
et
seront des fonctions symétriques des quatre côtés
et en se rappelant que
![{\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/0691eb4f57ea9b41f5cd096f727c8107e090e12f)
on aura
![{\displaystyle \operatorname {Sin} .(a,d)={\frac {2{\sqrt {MN}}}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}d}}\,;\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d2783282e7eddd7ae1689110c15cc793f08eeeb)
(III)
d’où l’on voit qu’ici les angles opposés ne sont pas supplément l’un de l’autre, comme dans le quadrilatère rectiligne inscrit.
En changeant respectivement
en
il vient
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .^{2}{\frac {1}{2}}(b,c)=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b-c-d)\operatorname {Cos} .{\frac {1}{4}}(a+c-b-d)\operatorname {Sin} .{\frac {1}{4}}(c+d+a-b)\operatorname {Sin} .{\frac {1}{4}}(d+a+b-c)}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}},\\\\&\operatorname {Cos} .^{2}{\frac {1}{2}}(b,c)=\\&{\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b+c+d)\operatorname {Cos} .{\frac {1}{4}}(b+c-a-d)\operatorname {Sin} .{\frac {1}{4}}(b+c+d-a)\operatorname {Sin} .{\frac {1}{4}}(a+b+c-d)}{\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}\,;\\\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4531a859d17c46fef6793a1e2131269b63592308)
mais on a