As the type of triangle is not mentioned in the question, I would take a right angled isosceles triangle right angled at B with #A(0,12), B(0,0) and C(12,0) #.
Now , the point D divides #AB# in the ratio #1:3#,
So, #D(x,y)=((m_1x_2+m_2x_1)/(m_1+m_2), (m_1y_2+m_2y_1)/(m_1+m_2))#
#=((1*0+3*0)/(1+3),(1*0+3*12)/(1+3))=(0,9)#
Similarly, #E(x,y)=((m_1x_2+m_2x_1)/(m_1+m_2), (m_1y_2+m_2y_1)/(m_1+m_2))#
#=((1*12+3*0)/(1+3),(1*0+3*0)/(1+3))=(9,0)#
Equation of line passing through #A(0,12) and E(3,0)# is
#rarry-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)#
#rarry-12=(0-12)/(3-0)(x-0)#
#rarr4x+y-12=0#.....[1]
Similarly, Equation of line passing through #C(12,0) and E(0,9)# is
#rarry-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)#
#rarry-0=(9-0)/(0-12)(x-12)#
#rarr3x+4y-36=0#.....[2]
Solving [1] and [2] by the rule of cross multiplication, we get,
#rarrx/(4xx(-2)-(-36)xx1)=y/(-3xx(-12)+4xx(-36)=)=1/(3-4*4)#
#rarrx=12/12 and y=108/13#
So, the co-ordinates of F is #(12/13,108/13)#.
Now, #(CF)^2/(FD)^2=((12/13-12)^2+(108/13-0)^2)/((0-12/13)^2+(9-108/13)^2)=(144^2+108^2)/(12^2+9^2)=144=12^2#
So, #(CF)/(FD)=12#
