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Question Number 194998 by a.lgnaoui last updated on 22/Jul/23

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

Commented by a.lgnaoui last updated on 22/Jul/23

Commented by a.lgnaoui last updated on 22/Jul/23

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