Estude! Com Prof. Caju

Em um triângulo isósceles ABC( AB = BC ). Se traça a altura AH e a bissetriz CM. Calcular BM se AC= 8 e m [tex3]\angle [/tex3]BAH = 3m [tex3]\angle [/tex3] HAC

[tex3]\mathsf{
\angle HAC = \alpha \implies \angle BAH = 3\alpha\\
\therefore \angle BAC = 3\alpha+\alpha = 4\alpha \implies \angle MCA = 2\alpha\\
\triangle HAC: 4\alpha+\alpha = 180^o \implies \alpha = 36^o \\
\triangle ACM: \angle AMC = 180 - 36-72 = 72^o \\
\therefore \triangle MCA_{(isoceles)} \implies C = CM = 8\\
\angle B = 180-2(72) = 36^o \\
\triangle CMB_{(isosceles)} \implies CM = \boxed{MB = 8}

}[/tex3]