1955 AMC 12 Problem 48

Place $A=(0,0),$ $B=(2,0),$ $C=(u,v)$ and compute the midpoints $D,E,F$

Use $\overrightarrow{FH}=\overrightarrow{AE}$, then locate $G$ where $BH$ reaches height $\frac{v}{2}$

Set $A=(0,0),$ $B=(2,0),$ $C=(u,v).$ Then $$\begin{aligned} D&=(1,0),\\ E&=\left(\frac{u+2}{2},\frac v2\right),\\ F&=\left(\frac u2,\frac v2\right). \end{aligned}$$ Since $\overrightarrow{FH}=\overrightarrow{AE},$ we get $H=(u+1,v).$ The horizontal line $FE$ meets $BH$ halfway from $B$ to $H,$ at $G=(\frac{u+3}{2},\frac{v}{2}).$ These coordinates verify that $AEHF$ is a parallelogram, $\overrightarrow{BH}=\overrightarrow{DC},$ $FG=\frac{3AB}{4},$ and $G$ is the midpoint of $BH,$ making $FG$ a median. But $$\begin{aligned} HE&=\frac12\sqrt{u^2+v^2},\\ HG&=\frac12\sqrt{(u-1)^2+v^2}, \end{aligned}$$ which are not generally equal.

Thus, statement B is not necessarily correct.