Estude! Com Prof. Caju

Problema Proposto
6 - Na figura se [tex3] \overset{\LARGE\frown}{AB} = 40^o [/tex3]calcular x.

Sendo B o centro do círculo menor.
B, G e F são colineares e BG=BD and FG=FE.
[tex3]\measuredangle DIE = 360 - 180 - 40 = 140^\circ\\
x=\measuredangle DGB+\measuredangle EGF\\\measuredangle ABF = 2\measuredangle DGB \implies \measuredangle DGB = \frac{\measuredangle ABF}{2}\\
\measuredangle AFB = 2\measuredangle EGF\implies \frac{AFB}{2}\\
\ \measuredangle ABF + \measuredangle AFB = 180^o-40^o = 140^o\\\therefore
x=\frac{1}{2}\left(\measuredangle ABF+\measuredangle AFB\right)=\frac{1}{2}\cdot140^{\circ}=\boxed{\color{red}70^{\circ}}.[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\measuredangle DIE = 360 - 180 - 40 = 140^\circ\\
x=\measuredangle DGB+\measuredangle EGF\\\measuredangle ABF = 2\measuredangle DGB \implies \measuredangle DGB = \frac{\measuredangle ABF}{2}\\
\measuredangle AFB = 2\measuredangle EGF\implies \frac{AFB}{2}\\
\ \measuredangle ABF + \measuredangle AFB = 180^o-40^o = 140^o\\\therefore
x=\frac{1}{2}\left(\measuredangle ABF+\measuredangle AFB\right)=\frac{1}{2}\cdot140^{\circ}=\boxed{\color{red}70^{\circ}}.[/tex3]

(Solução: Michael R.)