291
RÉSOLUES.
d’où, à cause de
, on conclura
![{\displaystyle \mathrm {A=B=C,} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/576377c30bf241b03aa7d24f49f6bc1ed0b97b2e)
comme ci-dessus.
2.o Pour le quadrilatère, on trouvera, en faisant absolument le même calcul,
![{\displaystyle a=\mathrm {\tfrac {\lambda \operatorname {Cos} .B}{\operatorname {Sin} .(A+B)}} ,\quad b=\mathrm {\tfrac {\lambda \operatorname {Cos} .C}{\operatorname {Sin} .(B+C)}} ,\quad c=\mathrm {\tfrac {\lambda \operatorname {Cos} .D}{\operatorname {Sin} .(C+D)}} ,\quad d=\mathrm {\tfrac {\lambda \operatorname {Cos} .A}{\operatorname {Sin} .(D+A)}} \ ;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94e16b2d92a8b890f980942bdadaf6dc3f889296)
![{\displaystyle a=\mathrm {\tfrac {\lambda \operatorname {Cos} .C}{\operatorname {Sin} .(C+D)}} ,\quad b=\mathrm {\tfrac {\lambda \operatorname {Cos} .D}{\operatorname {Sin} .(D+A)}} ,\quad c=\mathrm {\tfrac {\lambda \operatorname {Cos} .A}{\operatorname {Sin} .(A+B)}} ,\quad d=\mathrm {\tfrac {\lambda \operatorname {Cos} .B}{\operatorname {Sin} .(B+C)}} \ ;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9414926aaa7f3cab12e5266939f98880b68d37cd)
d’où on conclura :
![{\displaystyle \operatorname {Cos} .\mathrm {B} \operatorname {Sin} .\mathrm {(C+D)} =\operatorname {Cos} .\mathrm {C} \operatorname {Sin} .\mathrm {(A+B)} ,\quad \operatorname {Cos} .\mathrm {C} \operatorname {Sin} .\mathrm {(D+A)} =\operatorname {Cos} .\mathrm {D} \operatorname {Sin} .\mathrm {(B+C)} ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/306a3504e9cfe5b4d9912769460d73a506e75b80)
![{\displaystyle \operatorname {Cos} .\mathrm {D} \operatorname {Sin} .\mathrm {(A+B)} =\operatorname {Cos} .\mathrm {A} \operatorname {Sin} .\mathrm {(C+D)} ,\quad \operatorname {Cos} .\mathrm {A} \operatorname {Sin} .\mathrm {(B+C)} =\operatorname {Cos} .\mathrm {B} \operatorname {Sin} .\mathrm {(D+A)} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ceae127fdcc31938d22e82b22464131f850fbcb)
La condition
![{\displaystyle \mathrm {A+B+C+D=360^{\circ }} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3fbb322c3c19f6dc19d19e508768dfcb8eadfcba)
donne d’ailleurs
![{\displaystyle \operatorname {Sin} .\mathrm {(C+D)} =-\operatorname {Sin} .\mathrm {(A+B)} ,\qquad \operatorname {Sin} .\mathrm {(D+A)} =-\operatorname {Sin} .\mathrm {(B+C)} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/597bb9343ccc2d3e70a0038d1b373e8a8a4e0002)
substituant donc, il viendra
![{\displaystyle \operatorname {Cos} .\mathrm {B} =-\operatorname {Cos} .\mathrm {C} ,\ \operatorname {Cos} .\mathrm {C} =-\operatorname {Cos} .\mathrm {D} ,\ \operatorname {Cos} .\mathrm {D} =-\operatorname {Cos} .\mathrm {A} ,\ \operatorname {Cos} .\mathrm {A} =-\operatorname {Cos} .\mathrm {B} ~;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1531bad254283bc47405fc9ba53337bdaa5920f6)
et encore
![{\displaystyle \mathrm {A=C,\qquad B=D\ ;} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5dfa90fa13ea64d1a36ffae40b33d013210bb1d)
mêmes relations que ci-dessus.