In the given figure, D is the midpoint of side BC and AE

Solution:

(i)
In right-angled triangle AEC, applying Pythagoras theorem, we have:

$A C^{2}=A E^{2}+E C^{2}$

$\Rightarrow b^{2}=h^{2}+\left(x+\frac{a}{2}\right)^{2}=h^{2}+x^{2}+\frac{a^{2}}{4}+a x \ldots(i)$

In right - angled triangle AED, we have:

$A D^{2}=A E^{2}+E D^{2}$

$\Rightarrow p^{2}=h^{2}+x^{2} \quad \ldots(i i)$

Therefore,

from $(\mathrm{i})$ and $(\mathrm{ii})$

$b^{2}=p^{2}+a x+\frac{a^{2}}{4}$

(ii)
In right-angled triangle AEB, applying Pythagoras theorem, we have:

$A B^{2}=A E^{2}+E B^{2}$

$\Rightarrow c^{2}=h^{2}+\left(\frac{a}{2}-x\right)^{2} \quad\left(\because B D=\frac{a}{2}\right.$ and $\left.B E=B D-x\right)$

$\Rightarrow c^{2}=h^{2}+x^{2}-a x+\frac{a^{2}}{4} \quad\left(\because h^{2}+x^{2}=p^{2}\right)$

$\Rightarrow c^{2}=p^{2}-a x+\frac{a^{2}}{4}$

(iii)

Adding (i) and (ii), we get:

$\Rightarrow b^{2}+c^{2}=p^{2}+a x+\frac{a^{2}}{4}+p^{2}-a x+\frac{a^{2}}{4}$

$=2 p^{2}+a x-a x+\frac{a^{2}+a^{2}}{4}$

$=2 p^{2}+\frac{a^{2}}{2}$

(iv)
Subtracting (ii) from (i), we get:

$b^{2}-c^{2}=p^{2}+a x+\frac{a^{2}}{4}-\left(p^{2}-a x+\frac{a^{2}}{4}\right)$

$=p^{2}-p^{2}+a x+a x+\frac{a^{2}}{4}-\frac{a^{2}}{4}$

$=2 a x$