來個 15 另解.
利用圓外一點到圓的兩切線段等長

設 \( \overline{AF} =x, \overline{BF} = y \)

則 \( x+y = c \) ( \(a,b,c \) 為角 A, B, C 的對邊長)
    \( x+ b = x + \) (C 到圓 \( O_3 \) 的切線段長) \( = y +a \)

兩式解聯立得 \( x = s -b,  y = s-a \)，其中  \( s = \frac{a+b+c}{2} \)

同理得 \( \overline{AE}, \overline{CE}, \overline{BD}, \overline{CD} \)

六線段長為 \( \overline{AF} = s-b = \frac52, \overline{AE} = s-c = \frac32\)
                   \( \overline{BF} = s-a = \frac72, \overline{BD} = s-c = \frac32\)
                   \( \overline{CE} = s-a = \frac72, \overline{CD} = s-b = \frac52\)

所求  \( \frac{\triangle DEF}{\triangle ABC}=\frac{\triangle AEF+\triangle BFD+\triangle CDE}{\triangle ABC}=1-\left(\frac{3}{10}\cdot\frac{5}{12}+\frac{7}{12}\cdot\frac{3}{8}+\frac{7}{10}\cdot\frac{5}{8}\right)=\frac{7}{32} \)

填充3. 是您誤會題意

題意只說甲先乙後，沒有輪流取球