Answer:
[tex]\dfrac{ah}{a+h}[/tex]
Step-by-step explanation:
Consider similar triangles BMC and KMN.
In these triangles,
[tex]\dfrac{BC}{KN}=\dfrac{MO}{MP},[/tex]
where O is the point of intersection of BC and MP.
Note that segment OP has the length the same as the side of the square ABCD. Hence,
[tex]\dfrac{BC}{a}=\dfrac{h-BC}{h},\\ \\BC\cdot h=a(h-BC),\\ \\BCh=ah-aBC,\\ \\BC(h+a)=ah,\\ \\BC=\dfrac{ah}{a+h}.[/tex]
