Exercise 2. Let $\mathrm{ABC}$ be a triangle, $\mathrm{O}$ a point inside this triangle. The line parallel to (BC) passing through $O$ intersects $[C A]$ at $D$, the line parallel to $(C A)$ passing through $O$ intersects $[A B]$ at $E$, and the line parallel to $(A B)$ passing through $O$ intersects $[\mathrm{BC}]$ at $\mathrm{F}$. What is the value of the following sum of ratios:

$$
\frac{B F}{B C}+\frac{A E}{A B}+\frac{C D}{A C} ?
$$
## Solution to Exercise 2 First Solution:

Let $|A B C|$ denote the area of triangle $A B C$. We observe that

$$
\frac{B F}{B C}=\frac{|A B F|}{|A B C|}=\frac{|A B O|}{|A B C|}
$$

because triangles $A B F$ and $A B C$ have the same height and bases $B F$ and $B C$, and $A B O$ and $A B F$ have the same base $A B$ and two parallel heights of the same length (if $O'$ and $F'$ are the feet of the heights, then $OO'F'F$ is a rectangle).

From this, we deduce that

$$
\frac{B F}{B C}+\frac{A E}{A B}+\frac{C D}{A C}=\frac{|A B O|}{|A B C|}+\frac{|A O C|}{|A B C|}+\frac{|O C B|}{|A B C|}=1
$$