11. In the triangle $A B C, A B=A C=1, D$ and $E$ are the midpoints of $A B$ and $A C$ respectively. Let $P$ be a point on $D E$ and let the extensions of $B P$ and $C P$ meet the sides $A C$ and $A B$ at $G$ and $F$ respectively. Find the value of $\frac{1}{B F}+\frac{1}{C G}$.
11. Ans: 3

Let $D P=x, P E=y$, and $B C=a$. As $\triangle F D P$ is similar to $\triangle F B C$, we have $D F / B F=$ $x / a$. Thus $(B F-1 / 2) / B F=x / a$. Solving for $B F$, we have $1 / B F=2(a-x) / a$.

Similarly, from the fact that $\triangle G P E$ is similar to $\triangle G B C$, we obtain $1 / C G=$ $2(a-y) / a$. Consequently,
$$
\frac{1}{B F}+\frac{1}{C G}=(4 a-2(x+y)) / a=(4 a-a) / a=3 .
$$