For reference (by geometry): The angle $C$ of a triangle $ABC$ measures $53^°$. If you draw :$ BH \perp AC, HL \perp BC$ an $LQ \perp AB$; and If $QL \cap BH = M$, calculate $MH$, if $AC= 25$.(Answer:$12$)
My progresss: I made the drawing and used angle chasing..
$\triangle BHC \sim \triangle HLC \implies \frac{HB}{HL}=\frac{HC}{LC}=\frac{BC}{HC}\\
\frac{HB}{4k}=\frac{5k}{3k}=\frac{BC}{5k}=\frac{5}{5}\implies BC = \frac{25k}{3}\\
\triangle BLH \sim \triangle HLC \implies \frac{HL}{LC}=\frac{BL}{LH}=\frac{BH}{HC}\implies \\\frac{4k}{3k}=\frac{BL}{4k}=\frac{BH}{5k} \implies BL = \frac{16k}{3},5BL = 4BH\implies BH = \frac{20k}{3}\\
$
but some detail is missing
As you indicated in the diagram, $\angle HLM = \angle B$ and $\angle MHL = \angle C = 53^\circ$.
Therefore, $\triangle MLH \sim \triangle ABC$
Also using your work, $BC = \dfrac{25k}{3}$
$ \displaystyle \frac{MH}{AC} = \frac{HL}{BC} = \frac{4k}{25k/3} = \frac{12}{25}$
Given $AC = 25, MH = 12$