What's the measure of the segment $MN$ in the question below?

Question

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

Answer 1

Through B draw BX // AK. Through H, draw HY // AB.

If BX meets HY at Z, then ABZH is a parallelogram with BZ = AH = 14. Since N is the midpoint of the diagonal BH, AZ will go through N such that AN = NZ.

Applying the midpoint theorem to the purple triangle, we have $NM = 0.5CZ = 0.5\sqrt (BZ^2 + BC^2)$

Answer 2

$$MN=\sqrt{\left(\vec{MN}\right)^2}=\sqrt{\left(\frac{1}{2}\left(\vec{AH}+\vec{CB}\right)\right)^2}=\sqrt{\frac{1}{4}(14^2+48^2)}=25.$$

Answer 3

$\triangle AFH \sim \triangle BFC$

$FC = \frac{48}{14} HF, BF = \frac{48}{14} AF$

$2 FN = BF+HF = \frac{24}{7} AF + HF \tag1$

$\triangle ABF \sim \triangle HCF$

$\frac{CF}{HF} = \frac{BF}{AF} \implies CF = \frac{48}{14} HF$

$2 FM = CF - AF = \frac{24}{7} HF - AF \tag2$

Squaring $1$ and $2$ and adding,

$4 (FN^2 + FM^2) = \left(\frac{25}{7}\right)^2 (AF^2 + HF^2) = \left(\frac{25}{7}\right)^2 \cdot 14^2$

$MN^2 = 25^2 \implies MN = 25$