What’s the measure of the segment $MN$ in the question below

euclidean-geometryplane-geometrytrianglesvectors

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

enter image description here

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

Best Answer

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

enter image description here

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 $NM = 0.5CZ = 0.5\sqrt (BZ^2 + BC^2)$

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