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Resolution du probldme pose par sonukgindia (16.7.2023) voir Q194819 △ABC AM=AN=ADcos (𝛂/2) { ((AC=AM+MC=17 (1))),((AB=AN+NB =18 (2))) :} AB−AC=1=NB−MC (3) △CDE cos (C/2)=((CM)/(CD))=((CE)/(CD)) ⇒CM=CE △BDE cos (B/2)=((BE)/(BD))=((BN)/(BD))⇒BN=BE ⇒BE−CE=1 BC=BE+CE=2BE−1 △BEF BF=9 cos B=((BE)/9) BE=9cos B ⇒BC=18cos B−1 Posons BC=x x=18cos B−1 d apres triangle ABC AC^2 =AB^2 +BC^2 −2AB.BCcos B ⇒17^2 =18^2 +x^2 −36(((x+1)/(18))) x^2 +2x−35=0 alors x=5

$Resolution du probldme pose par$ $sonukgindia (16.7.2023)$ $voir Q 194819$ $△ ABC AM = AN = AD \cos \frac{α}{2}$ ${\begin{cases} AC = AM + MC = 17 (1) \\ AB = AN + NB = 18 (2) \end{cases}$ $AB - AC = 1 = NB - MC (3)$ $△ CDE \cos \frac{C}{2} = \frac{CM}{CD} = \frac{CE}{CD} \Rightarrow CM = CE$ $△ BDE \cos \frac{B}{2} = \frac{BE}{BD} = \frac{BN}{BD} \Rightarrow BN = BE$ $\Rightarrow BE - CE = 1$ $BC = BE + CE = 2 BE - 1$ $△ BEF BF = 9 \cos B = \frac{BE}{9}$ $BE = 9 \cos B \Rightarrow BC = 18 \cos B - 1$ $Posons BC = x x = 18 \cos B - 1$ $d apres triangle ABC$ ${AC}^{2} = {AB}^{2} + {BC}^{2} - 2 AB . BC \cos B$ $\Rightarrow 17^{2} = 18^{2} + x^{2} - 36 (\frac{x + 1}{18})$ $x^{2} + 2 x - 35 = 0$ $alors x = 5$

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soit A(2,1) B(3,2) C(4,3) points du plan( ox,oy) 1)Determiner l ′ equation du cercle qui passe par A; B; C ? 2) points d intersection du cercle avec l axe(ox,oy)?

$soit A (2, 1) B (3, 2) C (4, 3) points du$ $plan (ox, oy)$ $1) Determiner l^{'} equation du cercle qui$ $passe par A; B; C ?$ $2) points d intersection du cercle avec$ $l axe (ox, oy) ?$