Estude! Com Prof. Caju

São dados os ângulos consecutivos AOB, BOC e COD e são traçadas as bissetrizes OM
do ângulo AOB, ON do ângulo COD e OP do ângulo MON.
Se [tex3]m\angle MOC + m\angle MOD = 60^o + 4 m\angle BOP [/tex3], calcular a [tex3]m\angle AOB[/tex3]

[tex3]m\angle AOB = 2x \implies m\angle MOB = x\\
m\angle BOC = y\\
m\angle COD = 2z \implies m\angle CON = z[/tex3]
[tex3]
m\angle MOC = x + y\\
m\angle MOD = x + y + 2z\\
m\angle MON = x + y + z \implies m\angle MOP = \frac{x+y+z}{2}
[/tex3]
[tex3]m\angle BOP = |m\angle MOP - m\angle MOB| = \left|\frac{x+y+z}{2} - x\right| = \left|\frac{y+z-x}{2}\right| \\
= \frac{y+z-x}{2}[/tex3]
[tex3]m\angle MOC + m\angle MOD = 60^\circ + 4 \cdot m\angle BOP\\
( x+y) + (x+y+2z) = 60^\circ + 4 \cdot \left(\frac{y+z-x}{2}\right)\\
2x + 2y + 2z = 60^\circ + 2(y+z-x)\\
2x + 2y + 2z = 60^\circ + 2y + 2z - 2x\\
2x = 60^\circ - 2x\\
4x = 60^\circ\\
x = 15^\circ\\

m\angle AOB = 2x = 2 \cdot 15^\circ =\boxed{ 30^\circ}[/tex3]