
1. Let $A B C$ be a triangle and $D$ be a point on the segment $B C$ such that $D C=2 B D$. Let $E$ be the mid-point of $A C$. Let $A D$ and $B E$ intersect in $P$. Determine the ratios $B P / P E$ and $A P / P D$.

![](https://cdn.mathpix.com/cropped/2024_06_05_28688c797dda1f3c80e5g-1.jpg?height=323&width=507&top_left_y=641&top_left_x=424)


Solution: Let $F$ be the midpoint of $D C$, so that $D, F$ are points of trisection of $B C$. Now in triangle $C A D, F$ is the mid-point of $C D$ and $E$ is that of $C A$. Hence $C F / F D=$ $1=C E / E A$. Thus $E F \| A D$. Hence we find that $E F \| P D$. Hence $B P / P E=B D / D F$. But $B D=D F$. We obtain $B P / P E=1$.

In triangle $A C D$, since $E F \| A D$ we get $E F / A D=C F / C D=1 / 2$. Thus $A D=2 E F$. But $P D / E F=B D / B F=1 / 2$. Hence $E F=2 P D$. Therefore This gives

$$
A P=A D-P D=3 P D
$$

We obtain $A P / P D=3$.

(Coordinate geometry proof is also possible.)
