What's the measure of the segment $MN$ in the question below? For reference:(exact copy of the question) In the triangle $ABC$, $H$ is the orthocenter, $M$ and $N$ are midpoints of $AC$ and $BH$ respectively.
Calculate $MN$, if $AH=14$ and $BC=48$ (answer: $25$)
My progress..my drawing according to the statement and the relationships I found

we have several similarities
$\triangle AKC \sim \triangle BKH\\
\triangle AHE \sim BHK \sim \triangle BCE\implies\\
\frac{14}{48} = \frac{HE}{CE}=\frac{AE}{BE}\\\frac{BH}{48} = \frac{KH}{CE}=\frac{BK}{BE}\\
\frac{14}{BH} = \frac{HE}{HK}=\frac{AE}{BK}\\
\triangle MKC \sim \triangle NKH\\
\triangle NBK \sim \triangle MAK$

$\triangle AFH \sim \triangle BFC$
$FC = \frac{48}{14} HF, BF = \frac{48}{14} AF$
$2 FN = BF+HF = \frac{24}{7} AF + HF \tag1$
$\triangle ABF \sim \triangle HCF$
$\frac{CF}{HF} = \frac{BF}{AF} \implies CF = \frac{48}{14} HF$
$2 FM = CF - AF = \frac{24}{7} HF - AF \tag2$
Squaring $1$ and $2$ and adding,
$4 (FN^2 + FM^2) = \left(\frac{25}{7}\right)^2 (AF^2 + HF^2) = \left(\frac{25}{7}\right)^2 \cdot 14^2$
$MN^2 = 25^2 \implies MN = 25$
