Задачаа

[Гост]

Screenshot_20240303_122532_Drive.jpg (60.54 KiB) Прегледано 2283 пъти

[ammornil]

[tex]s_{_{AC}}\cap s_{_{BC}}=O \Rightarrow OA=OB=OC=R_{k} \rightarrow k(O; R_{k}) \\ s_{_{AC}}\cap k=D, s_{_{BC}}\cap k=E \\ M \in s_{_{AC}} \Rightarrow AM=MC \Rightarrow \angle{MAC}=\angle{ACM}=\varphi \\ N \in s_{_{BC}} \Rightarrow BN=NC \Rightarrow \angle{NBC}=\angle{BCN}=\mu \\ \triangle{ABC} \rightarrow \varphi+\varphi+60^{\circ}+\mu+\mu=180^{\circ} \Rightarrow \varphi+\mu=60^{\circ} \\ \begin{cases} \widehat{ADC} = 2\cdot{\mu} \\ \widehat{CEB} = 2\cdot{\varphi} \end{cases} \Rightarrow \widehat{ADB} = \widehat{ADC}+\widehat{CEB} =2\cdot{\varphi}+ 2\cdot{\mu}=120^{\circ} \\ \angle{AOB}=\widehat{ADB}=120^{\circ} \\ \triangle{AOB} \rightarrow OA=OB \Rightarrow \angle{OAB}=\angle{OBA}=x \Rightarrow x+x+120^{\circ}=180^{\circ} \Rightarrow x=30^{\circ} \\ \boxed{\angle{MAO}=x=30^{\circ}} \\[/tex]Частта от симетралата на [tex]AC[/tex] между точка [tex]M[/tex] и страната [tex]AC[/tex] е височина, медиана и ъглополовяща в [tex]\triangle{AMC} \Rightarrow \angle{AMD}=\angle{CMD} \Rightarrow \widehat{AD}=\widehat{CD}=\mu \\[/tex] Аналогично [tex]\triangle{BNC} \Rightarrow \angle{CNE}=\angle{BNE} \Rightarrow \widehat{CE}=\widehat{EB}=\varphi \\ \widehat{DCE}=\widehat{DC}+\widehat{CE}=\mu + \varphi = 60^{\circ} \\ \boxed{\angle{MON}=\widehat{DCE}=60^{\circ}}\\[/tex]

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