What's the distance from the vertex C to the segment NL in the question below?

Question

For reference(exact copy of the question): In an acute triangle ABC the heights $AN, ~BL~ and~ CM$ are drawn up. If $\measuredangle C = 45^o°$ and CM = 6; find the distance from the vertex C to NL.(answer: $3\sqrt2$)

My progres..I found an orthic triangle $MNL$ and a cyclic quadrilateral $HNLC$

$\measuredangle LHN = 135^o$

$\measuredangle LBC = 45^o \implies \triangle HBN (isosceles)$

my drawing

Answer 1

I refer to internal angles of triangles $\triangle ABC$ as $\angle A, \angle B$ and $ \angle C$.

First observe that $ABNL$ is cyclic and we establish that,
$\angle BLN = \angle BAN = 90^\circ - \angle B ~ $ and $ ~ \angle CLN = \angle B$

So, $\triangle CDL \sim \triangle CMB$. It follows that $ \frac{CD}{CL} = \frac{CM}{ CB}$

Also note that $CL = \frac{CB}{\sqrt2}$

$ \therefore CD = \frac{CM}{\sqrt2} = 3 \sqrt2$