Nível I ⇒ Problema 37 - Triângulos -Vol. 2 Tópico resolvido

[petrasMOD]

Em um triângulo ABC se traça a bissetriz interna AM do ângulo A(M em BC), depois se traça MN || AC, (N em AB) e a bissetriz ND do ângulo MNA (D em AC).
Se [tex3]m \measuredangle ABC - m\measuredangle ACB = 80^o\\
m \measuredangle MDN = m\measuredangle BMD[/tex3]
Calcule [tex3]m \measuredangle NDM[/tex3].
A) 50^o
B) 60^o
C) 70^o
D) 80^o
E) 75^o

Resposta

---

Resposta: C

[petrasMOD]

[tex3]\mathsf{
\angle MAB \cong \angle DAM = \alpha\\
\angle B = \beta\\
\angle C =\theta\\
\beta - \theta =80^o (i)\\
\triangle ABC : 2\alpha+\beta+\theta = 180^o (ii)\\
De(i)e(ii): \alpha +\beta = 50^o(iii) \\
MN \parallel AC \implies \angle MNB = 2 \alpha:\angle BMN = \beta\\
\angle ANM = 180^o -2\alpha \implies \angle NMA = \alpha\\
\therefore \triangle ANM_{(isosc)} \implies ND \perp AM
\therefore ANMD_{(losang.)}\\
\angle MDN = 90-\alpha(iv) \\
\angle BMD = 2\alpha + \beta(v)\\
(iv)=(v): 90-\alpha = 2\alpha +\beta \implies 3\alpha + \beta = 90^o (vi)\\
De(iii)e(vi):\alpha = 20^o : \beta = 30^o\\
\therefore \boxed{\angle NDM\cong \angle BMD = 2\alpha+\beta = 2.20^o+30^o = 70^o}


}[/tex3]