17.48 Given triangle $A B C$. Median $A E, B F$, $C D, F H$ is parallel and equal to $A E$, connect $B H$ and $H E$, extend $F E$ to intersect $B H$ at $G$. Which of the following statements is not necessarily true?
(A) $A E H F$ is a parallelogram.
(B) $H E=E G$.
(C) $B H=D C$.
(D) $F G=\frac{3}{4} A B$.
(E) $F G$ is the median of triangle $B F H$.
(6th American High School Mathematics Examination, 1955)
[Solution] (1) It is easy to prove that $A E \underline{=} F H$, so (A) is true;
(2) From (A), we get $E H=A F=F C$, and $E H / / F C$, hence quadrilateral $C F E H$ is a parallelogram.
$\therefore C H / / E F / / A B$, and $C H=E F=B D$.
$\therefore$ Quadrilateral $C D B H$ is a parallelogram. Hence (C) is true;
(3) By the given, $E F=\frac{1}{2} A B$. Connecting $E D$, and it can be proven that $H E=E D$, and they are collinear $(E H / / A F, D E / / A F)$.
Thus, $E G=\frac{1}{2} B D$,

So $F G=E F+E G=\frac{1}{2} A B+\frac{1}{2} B D=\frac{3}{4} A B$
Hence (D) is true;
(4) Since $G$ is the midpoint of $B H$, then (E) is true;
(5) If $H E=E G$ holds, it requires $D H=D B$, but $\triangle A B C$ is an arbitrary triangle, so $H E=E G$ does not necessarily hold.
In summary, the answer is (B).