Write $AS$ as a multiple of $AD$ and $CS$ as a multiple of $CE$.

Question

In a triangle $ABC$ the point $D$ is on $BC$ such that $\overline{BD}=\frac{2}{3}\cdot \overline{BC}$ and the point $E$ is on $AB$ such that $\overline{AE}=\frac{3}{5}\cdot \overline{AB}$.

The lines $AD$ and $CE$ intersect in $S$.

Write $AS$ as a multiple of $AD$ and $CS$ as a multiple of $CE$.

We have a triangle as follows, right?

But how can we get an information for $AS$ and $CS$ ? Could you give me a hint? Do we maybe use vector operations?

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EDIT :

Using vectors we have $$\vec{AS}+\vec{SD}=\vec{AD} \Rightarrow \vec{AS}=\vec{AD}-\vec{SD} \Rightarrow \vec{AS}=\vec{AD}+\vec{DS}$$ Does this help us ? How can we continue?

Answer 1

Hint:

Applying Ceva's theorem we have:

$$\frac{AE}{EB}\cdot\frac{BD}{DC}\cdot\frac{CG}{AG}=1$$

Puting values we obtain:

$AG=3 CG$

As can be seen in figure, M is midpoint of BC(you have to prove it) and in triangle ABD; AM, DE and BC intersect in point F. Also we have:

$MD=\frac 23 BC-\frac 12 BC=\frac 16 BC$

Applying Ceva's theorem on this triangle we have:

$$\frac{AE}{EB}\cdot\frac{BM}{MD}\cdot \frac{DS}{AS}=\frac 32\cdot \frac {BC/2}{BC/6}\cdot \frac{DS}{AS}=1$$

Which gives $AS=4.5 DS$ , or $AS=\frac 9{11}AD$

Similarly you can find other ratio.

Answer 2

Here is another solution. I solve the problem for $AS$ and $AD$. You can do the same for $CS$ and $CE$.

As shown in the figure below, from $A$ , $B$ and $D$ draw $AH$ , $BK$ and $DL$ perpendicular to $CE$ . We have: $$\frac{AH}{BK} = \frac{AE}{BE} = \frac 32$$ $$\frac{BK}{DL} = \frac{BC}{DC} = \frac 31$$ $$\therefore \frac{AH}{DL} = \frac 32 \times \frac 31 = \frac 92 $$ Now we can calculate $\frac{AS}{AD}$: $$\frac{AS}{SD} = \frac{AH}{DL} = \frac 92 $$ $$\therefore \frac{AS}{AD} = \frac{AS}{AS+SD} = \frac{9}{9+2} = \frac{9}{11} $$

Answer 3

Here is a method using vectors.

Let position vectors of the vertices be $A,B,C$. Then $$D=\frac{B+2C}{3} \quad ; \quad E=\frac{2A+3B}{5}$$

Let $AS=\lambda AD$ and $CS=\mu CE$, where $\lambda, \mu$ are to be found.

Then $S$ divides $AD$ in ratio $\lambda : (1-\lambda)$ and $CE$ in ratio $\mu : (1-\mu)$, we can write position vector of $S$ in two ways as

$$S=\lambda D+(1-\lambda)A=(1-\mu)C+\mu E$$ $$\Rightarrow (1-\lambda)A+\frac{\lambda}{3}B+\frac{2\lambda}{3}C=\frac{2\mu}{5}A+\frac{3\mu}{5}B+(1-\mu)C$$

This can be solved for $\lambda, \mu$ by equating the coefficients.