Find $\frac{FE}{FD}$ in a triangle where you know $AB$, $AE$, $EC$

Question

In the triangle $ABC$, $AB=39$, $AE=9$, $EC=16$ ($E$ is a point on $AC$). Find the ratio $\frac{FE}{FD}$.

So far, using the right triangle altitude theorem and Pitagora's theorem, I have found:

$DE=12\\AD=15\\DC=20\\BD=36$

But I don't know how to continue it. Can you help me? Thanks!

Answer 1

Let us make a slightly more accurate picture...

I've also constructed the projection $X$ of $B$ on the line $AC$.

Then $$ \begin{aligned} \frac{16}{EX} &= \frac{CE}{EX} = \frac{CD}{DB} = \frac{20}{36} \ , \\[4mm] \frac{12}{BX} &= \frac{DE}{BX} = \frac{CD}{CB} = \frac{20}{56} \ , \\[4mm] &\qquad\text{ and from here} \\[4mm] EX&= \frac{144}5=28.8\ ,\\ BX&= \frac{168}5=33.6\ , \\[4mm] \frac{FE}{FA} &= \cot \widehat {AEF} = \cot \widehat {XEB} \\ &= \frac{XE}{XB}= \frac{144}{168}= \frac 67\ , \\[4mm] \frac{FA}{FD} &= \frac {\sin \angle ADF} {\sin \angle DAF} = \frac {\sin \angle ABF} {\sin \angle DBF} = \frac {\sin \angle ABE} {\sin \angle EBC} \\ & = \frac {2\text{Area}(ABE)/(BA\cdot BE)} {2\text{Area}(EBC)/(BC\cdot BE)} = \frac {\text{Area}(ABE)/BA} {\text{Area}(EBC)/BC} = \frac {AE/BA} {EC/BC} \\ &=\frac{9/39}{16/56}=\frac{21}{26}\ , \\[4mm] \frac{FE}{FD} &= \frac{FE}{FA} \cdot \frac{FA}{FD} \\ &=\frac 67\cdot \frac{21}{26}=\frac 9{13}\ . \end{aligned} $$ (I have to submit, check the computation later, but the computational bridge should work.)

Answer 2

Hint:

This problem is screaming for coordinate system (say with $BC$ as x-axis and $AD$ as y-axis.)