$\triangle ABC$ is a right triangle with $\angle ACB=90^\circ$. Let $CM=m_c$ be the median and $CH=h_c$ be the altitude through $C$.
I want to understand how I should approach problems with similar configuration. In what order are the points on $AB$? That depends on the angles of the triangle, doesn't it? I want to see it myself that if $\measuredangle BAC>\measuredangle ABC$, then $AH<AM$. So if we suppose that $CH'\perp AB, AH'>AM$, how to show this is not possible? I have not studied trigonometry!!
In the standard notation $$AM=m_c=\frac{\sqrt{a^2+b^2}}{2}.$$ Also, $$AC^2=AH\cdot AB,$$ which gives $$AH=\frac{b^2}{\sqrt{a^2+b^2}}.$$ Thus, $$AM-AH=\frac{\sqrt{a^2+b^2}}{2}-\frac{b^2}{\sqrt{a^2+b^2}}=\frac{a^2-b^2}{2\sqrt{a^2+b^2}}.$$ Can you end it now?