Cho tam giác ABC cân ( AB = AC ) với góc A = 45 độ, nội tiếp trong đường tròn tâm O

$$\begin{gathered} \mathrm{Bài} 4:\mathrm{Cho} \mathrm{tam} \mathrm{giác} \mathrm{cân} \mathrm{ABC} (\mathrm{AB}=\mathrm{AC}) \mathrm{với} \mathrm{góc} \mathrm{A} ={45}^{°},\mathrm{nội} \mathrm{tiếp} \mathrm{trong} \\ \mathrm{đường} \mathrm{tròn} \mathrm{tâm} \mathrm{O}.\mathrm{Đường} \mathrm{tròn} \mathrm{đường} \mathrm{kính} \mathrm{BC} \mathrm{cắt} \mathrm{AB} \mathrm{ở} \mathrm{E} ,\mathrm{cắt} \mathrm{AC} \mathrm{ở} \mathrm{F}. \end{gathered}$$
$$\begin{gathered} \mathbf{a}\mathbf{,}\mathbf{Cmr}\mathbf{:}\mathbf{O}\mathbf{ }\mathbf{thuộc}\mathbf{ }\mathbf{đường}\mathbf{ }\mathbf{tròn}\mathbf{ }\mathbf{đường}\mathbf{ }\mathbf{kính}\mathbf{ }\mathbf{BC}\mathbf{.} \\ \mathbf{b}\mathbf{,}\mathbf{Cm}\mathbf{ }\mathbf{tam}\mathbf{ }\mathbf{giác}\mathbf{ }\mathbf{AEC}\mathbf{,}\mathbf{AFB}\mathbf{ }\mathbf{là}\mathbf{ }\mathbf{những}\mathbf{ }\mathbf{tam}\mathbf{ }\mathbf{giác}\mathbf{ }\mathbf{vuông}\mathbf{ }\mathbf{cân}\mathbf{.} \\ \mathbf{c}\mathbf{,}\mathbf{Cm}\mathbf{ }\mathbf{tứ}\mathbf{ }\mathbf{giác}\mathbf{ }\mathbf{EOFB}\mathbf{ }\mathbf{là}\mathbf{ }\mathbf{hình}\mathbf{ }\mathbf{thang}\mathbf{ }\mathbf{cân}\mathbf{.}\mathbf{Suy}\mathbf{ }\mathbf{ra}\mathbf{ }\mathbf{EF}\mathbf{=}\mathbf{BC}\frac{\sqrt{\mathbf{2}}}{\mathbf{2}}\mathbf{.} \end{gathered}$$