Let $D$ and $E$ be points on $AB$ and $BC$ of $\Delta ABC$ such that $AD = 7BD$ , $BE = 10CE$ . Let $AE$ and $BC$ meet at point $F$

Question

Let $D$ and $E$ be points on segments of $AB$ and $BC$ of $\Delta ABC$ such that $AD = 7BD$ , $BE = 10CE$ . Let $AE$ and $BC$ meet at point $F$. If $AF = k \cdot FE$ , find $[k]$ .

What I Tried: Here is a picture :-

The first thing which I did was to try the process in Geogebra. There I got the value of $k$ to be approximately $7.74$ , so $[k] = 7,$ and I think that should be the answer. I also think that the initial value of $k$ varies, but $[k]$ is always $7$ . How will I be able to come up with a solution then?

Can anyone help me? Thank You.

Answer 1

Hint:Use Menelaus theorem in $\Delta ABE$ you will get result directly

$$\frac{BC}{CE}*\frac{EF}{FA}*\frac{AD}{DB}=1$$

About Menelaus https://artofproblemsolving.com/wiki/index.php/Menelaus%27_Theorem

Answer 2

Use mass points method.

Place masses $70$ at $B$, $10$ at $A$, $7$ at $C$.

This gives $77$ at $E$ implying

$$ \dfrac{AF}{FE}= 7.7$$

Answer 3

Say $\angle BCD = \theta$.

Using Sine Law in $\triangle BCD$,

$\frac{\sin (\angle BDC)}{11y} = \frac{\sin \theta}{x}$ ...(i)

In $\triangle CEF$,

$\frac{\sin (\angle EFC)}{10y} = \frac{\sin \theta}{EF}$ ..(ii)

In $\triangle ADF$,

$\frac{\sin (\angle AFD)}{7x} = \frac{\sin \angle ADC}{AF}$ ..(iii)

$\angle ADC = 180^0 - \angle BDC$ (same sine value) and $\angle EFC = \angle AFD$

Substituting and solving, we get $\frac{AF}{FE} = \frac{77}{10}$