What's the value of the $\frac{HP}{OL}$ in the triangle below?

Question

For reference:

In the triangle $ABC$; $M, N$ and $P$ are midpoints of $AB, BC$ and $AC$ respectively. If $H$ and $O$ are the orthocenter and circumcenter of this triangle and also $L$ is the midpoint of $MN$, calculate $\frac{HP}{OL}$ (Answer:2)

My progress:

$HB=2OP$(by property)

$MN = \frac{AC}{2}\implies LM = \frac{AC}{4} = LN\\\triangle HEQ \sim \triangle HDP \implies\\\frac{HE}{HD} = \frac{HQ}{HP} = \frac{EQ}{DP}\\\triangle BEM \sim \triangle BDA; k =\frac{1}{2}\\ \triangle BDC \sim \triangle BEN; k =\frac{1}{2}$

$LO$ is a median of $\triangle OMN$.

By geogebra $LO$ is parallel to $HP$.

Answer 1

$\angle HAC = 90^\circ - \angle C$. Also, $\angle ONM = 90^\circ - \angle C$. You can similarly check other angles to conclude,

$\triangle AHC \sim \triangle NOM$.

That leads to, $ ~ \displaystyle \frac{HP}{OL} = \frac{AC}{NM} = 2$

Answer 2

We can just calculate it.
Since $ \vec{HP}=(\vec{HA}+\vec{HC})/2=\vec{HO}+(\vec{OA}+\vec{OC})/2=\vec{BO}+\frac{1}{2}\vec{AO}+\frac{1}{2}\vec{CO} $
And $\vec{LO}=\vec{LB}+\vec{BO}=(\vec{MB}+\vec{NB})/2+\vec{BO}=(\vec{AB}+\vec{CB})/4+\vec{BO}=\frac{1}{2}\vec{BO}+\frac{1}{4}\vec{AO}+\frac{1}{4}\vec{CO} $
Hence $ \vec{HP}=2\vec{LO} $ as desired.