How to compute a particular sum of ratios in a triangle?

Question

In triangle $ABC,$ $AD$ is median and a secant $EF$ cuts $AD$ in the ratio $p:q$ and $G$ is fixed.

We wish to show that $BE/EA + CF/FA$ is constant and is equal to $2q/p.$

I tried in the following way.

Draw a line from $F$ and $E$ parallel to $BC$ and it meets $AB$ and $AC$ at $H$ and $I,$ respectively.

I take $KG= a$ and $GL=b.$ Then using similarity, $BE/EA + CF/FA=(q-b)/(p+b)+ (q+a)/(p-a).$

Again triangle $KGF$ and $EGL$ are similar and hence $a/b=(p-a)(q-b)/[(p+a)(q+b)].$ I am trying to prove it and will post an answer. If anyone can suggest any hint, it will be helpful. Thanks in advance.

Answer 1

Draw $BX, CY$ parallel to $EF$ intersecting $AD$ at $X,Y$. The two blue triangles are congruent. Can you continue from here?

Answer 2

Triangle BDX and CDY are congruent. So DX=DY=a (say). Now by the similarity between triangles AEG and ABX, we have $AE/EB=AG/GX=p/(q-a)$. By similarity of triangles AGF and AYC $AF/FC=AG/GY=p/(q+a)$.

Therefore $EB/AE+FC/AF=(q-a)/p+(q+a)/p=2q/p$.