May 23, 2018

As the type of triangle is not mentioned in the question, I would take a right angled isosceles triangle right angled at B with $A \left(0 , 12\right) , B \left(0 , 0\right) \mathmr{and} C \left(12 , 0\right)$.

Now , the point D divides $A B$ in the ratio $1 : 3$,

So, $D \left(x , y\right) = \left(\frac{{m}_{1} {x}_{2} + {m}_{2} {x}_{1}}{{m}_{1} + {m}_{2}} , \frac{{m}_{1} {y}_{2} + {m}_{2} {y}_{1}}{{m}_{1} + {m}_{2}}\right)$

$= \left(\frac{1 \cdot 0 + 3 \cdot 0}{1 + 3} , \frac{1 \cdot 0 + 3 \cdot 12}{1 + 3}\right) = \left(0 , 9\right)$

Similarly, $E \left(x , y\right) = \left(\frac{{m}_{1} {x}_{2} + {m}_{2} {x}_{1}}{{m}_{1} + {m}_{2}} , \frac{{m}_{1} {y}_{2} + {m}_{2} {y}_{1}}{{m}_{1} + {m}_{2}}\right)$

$= \left(\frac{1 \cdot 12 + 3 \cdot 0}{1 + 3} , \frac{1 \cdot 0 + 3 \cdot 0}{1 + 3}\right) = \left(9 , 0\right)$

Equation of line passing through $A \left(0 , 12\right) \mathmr{and} E \left(3 , 0\right)$ is

$\rightarrow y - {y}_{1} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} \left(x - {x}_{1}\right)$

$\rightarrow y - 12 = \frac{0 - 12}{3 - 0} \left(x - 0\right)$

$\rightarrow 4 x + y - 12 = 0$.....[1]

Similarly, Equation of line passing through $C \left(12 , 0\right) \mathmr{and} E \left(0 , 9\right)$ is

$\rightarrow y - {y}_{1} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} \left(x - {x}_{1}\right)$

$\rightarrow y - 0 = \frac{9 - 0}{0 - 12} \left(x - 12\right)$

$\rightarrow 3 x + 4 y - 36 = 0$.....[2]

Solving [1] and [2] by the rule of cross multiplication, we get,

$\rightarrow \frac{x}{4 \times \left(- 2\right) - \left(- 36\right) \times 1} = \frac{y}{- 3 \times \left(- 12\right) + 4 \times \left(- 36\right) =} = \frac{1}{3 - 4 \cdot 4}$

$\rightarrow x = \frac{12}{12} \mathmr{and} y = \frac{108}{13}$

So, the co-ordinates of F is $\left(\frac{12}{13} , \frac{108}{13}\right)$.

Now, ${\left(C F\right)}^{2} / {\left(F D\right)}^{2} = \frac{{\left(\frac{12}{13} - 12\right)}^{2} + {\left(\frac{108}{13} - 0\right)}^{2}}{{\left(0 - \frac{12}{13}\right)}^{2} + {\left(9 - \frac{108}{13}\right)}^{2}} = \frac{{144}^{2} + {108}^{2}}{{12}^{2} + {9}^{2}} = 144 = {12}^{2}$

So, $\frac{C F}{F D} = 12$