Задача с триъгълник

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[Nathi123]

Нека [tex]\angle CAB=\alpha ; \angle ABC = \beta \Rightarrow \angle MAB=\frac{\alpha}{2}+\frac{\alpha}{4}=\frac{3\alpha}{4}[/tex] Аналогично [tex]\angle MBA=\frac{3\beta}{4}[/tex]
[tex]\triangle AMB\Rightarrow \angle AMB=180^\circ - ( \angle MAB + \angle ABM) = 180^\circ -\frac{3}{4}(\alpha + \beta); \triangle ABC\Rightarrow \alpha + \beta = 140^\circ[/tex],защото [tex]\angle ACB =40^\circ[/tex]
[tex]\Rightarrow \angle AMB= 180^\circ -\frac{3.140}{4}=180^\circ -105^\circ =75^\circ[/tex].