Olhando a imagem, sabemos que :
AD é raio e AF também é raio, logo AD = AF ;
AC = AF + CF e CF = CB = a/2 ;
Daí, façamos pitagoras no triangulo ABC :
[tex]\displaystyle \sf AC^2 = AB^2 + CB^2 \\\\ AC^2 = a^2+\left(\frac{a}{2}\right)^2 \\\\\\ AC^2 = a^2+\frac{a^2}{4} \\\\ AC^2 = \frac{5a^2}{4} \\\\\ AC = \frac{a\sqrt{5}}{2}[/tex]
Dai :
[tex]\displaystyle \sf AC = AF + CF \\\\ \frac{a\sqrt{5}}{2}=AF+\frac{a}{2} \\\\\\ AF = \frac{a\sqrt{5}-a}{2} \to AF = \frac{a(\sqrt{5}-1)}{2} \\\\\\\ \underline{\text{Mas AF = AD. Portanto }}: \\\\ \huge\boxed{\sf AD = \frac{a(\sqrt{5}-1)}{2} \ }\checkmark[/tex]
letra d
