Finding ratio of cevian lines

Question

I am preparing for an exam and doing some pratice problems. So I'm having a difficult time with this problem. At first I thought the ratio was 2:1 and then I also thought I would be able to use the angle bisector theorem but then I realized that cevians could be altitudes, medians or angle bisectors. So theorems are not helpful with this problem.

Basically, I asked my teacher for help and he told me that the answer is

EF:FC is $10:7$

and DF:FA is $3:14$

I've been trying to find out how he got these ratios but i'm not getting anywhere. Any ideas?

Answer 1

A general method to solving these kinds of things is through the Ratio Lemma, which is often applied in high school olympiad geometry. It states that if $AD$ is a cevian in $\triangle ABC$, then $\dfrac{BD}{DC}=\dfrac{AB}{AC}\cdot\dfrac{\sin\angle BAD}{\sin\angle CAD}$. The proof of this is quite simple; just apply the sine law to triangles $ABD$ and $CAD$.

So for this problem, the ratios $\sin\angle BAD:\sin\angle CAD$ is clearly equal to $\sin\angle EAF:\sin\angle CAF$. So $EF:FC$ is equal to $(BD/DC)(AE/AB)=10/7$. $DF:FA$ is calculated similarly.

The Ratio Lemma can also be used to prove many basic facts in projective geometry; e.g. the invariance of the cross-ratio when projected through a point onto a line.

Answer 2

1. After joining BF, label the areas as shown. For example, p = area of triangle BEF = [BEF].

2. The rule:- “$\triangle$s having the same altitude, ratio of their areas is proportional to the ratio of their bases.”

For example, $\dfrac {p}{z} = \dfrac {3}{4}$. This means $p = \dfrac {3}{4}z$. Do the same to $\dfrac {q}{x} = …$ to get $q = ? x$. Therefore, $[BEFD] = p + q = \dfrac {3}{4}z + ?x$.

3. Apply the same rule to get $\dfrac {[ABD]}{[ADC]} = \dfrac {5}{2}$. Combining all the results, you should get $\dfrac {z}{y} = \dfrac {10}{7}$ (with x disappears miraculously).

4. Apply the converse of the rule to get $\dfrac {EF}{FC} = … $.

Answer 3

Menelaus' theorem for $\triangle BCE$ and transversal $DA$ gives $ \frac{EF}{FC}\; \frac{CD}{DB}\; \frac{BA}{AE} = 1 = \frac{EF}{FC}\; \frac{2}{5}\; \frac{3+4}{4}$ thus $\frac{EF}{FC} = \frac{10}{7}$. Permute $5 \leftrightarrows 3, 2 \leftrightarrows 4$ to get $\frac{DF}{FA} = \frac{3}{14}$.

Answer 4

Use mass points. We assign B a mass of 4. So A has a mass of 3 and C has a mass of 10. E has a mass of 7 and D has a mass of 14. So we have $$ \frac{AF}{FD}= \frac{14}{3}$$ and $$\frac{EF}{FC}= \frac{10}{7}$$.