Question

In a triangle $ABC,$ points $D$ and $E$ are on segments $BC$ and $AC$ such that $BD=3DC$ and $AE=4EC.$ Point $P$ is on the line $ED$ such that $D$ is the midpoint of segment $EP.$ Lines $AP$ and $BC$ intersect at point $S.$ Find the ratio $BS:SD$.

• A. $3:2$
• B. $5:2$
• C. $7:2$
• D. $9:2$

Answer 1

The correct option is C $7:2$

Let F denote the midpoint of the segment AE.
Then it follows that DF is parallel to AP.
$\therefore$ in triangle ASC we have $\frac{CD}{SD}=\frac{CF}{FA}=1.5$.
But $DC=\frac{BD}{3}=\frac{(BS+SD)}{3}.$
$\therefore \frac{BS}{SD}=\frac{7}{2}$