In the given figure, D is the mid-point of side BC and AE ⊥ BC

Solution:

(i) It is given that $\mathrm{D}$ is the midpoint of $\mathrm{BC}$ and $B C=a$.

Therefore, $B D=D C=\frac{a}{2}$...(1)

Using Pythagoras theorem in the right angled triangle AED,

$A D^{2}=A E^{2}+E D^{2}$....(2)

Let us substitute $A D=p, A E=h$ and $E D=x$ in equation (2), we get

$p^{2}=h^{2}+x^{2}$

Let us take another right angled triangle that is triangle AEC.

Using Pythagoras theorem,

$A C^{2}=A E^{2}+E C^{2}$...$.(3)$

Let us substitute $A E=h$ and $E C=x+\frac{a}{2}$ in equation (3) we get,

Here we know that $D C=\frac{a}{2}$ and $E D=x$.

$E C=E D+D C$

Substituting $A C=b, D C=\frac{a}{2}$ and $E D=x$ we get $E C=\left(x+\frac{a}{2}\right)$

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

      $b^{2}=h^{2}+x^{2}+x a+\frac{a^{2}}{4}$....(4)

From equation (1) we can substitute $h^{2}+x^{2}=p^{2}$ in equation (4),

$b^{2}=h^{2}+x^{2}+x a+\frac{a^{2}}{4}$...(5)

(ii) Using Pythagoras theorem in right angled triangle AEB,

$A B^{2}=A E^{2}+B E^{2}$....$(6)$

 

We know that AB = c and AE = h now we will find BE.

$B D=B E+E D$

Therefore, $B E=B D-E D$

We know that $B D=\frac{a}{2}$ and $E D=x$ substituting these values in $B E=B D-E D$ we get,

$B E=\frac{a}{2}-x$

Now we will substitute $\mathrm{AB}=\mathrm{c}, \mathrm{AE}=\mathrm{h}$ and $B E=\frac{a}{2}-x$ in equation (6) we get,

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

$c^{2}=h^{2}+\frac{a^{2}}{4}-a x+x^{2}$....(7)

Let us rewrite the equation (7) as below,

$c^{2}=h^{2}+x^{2}+\frac{a^{2}}{4}-a x$.....(8)

From equation (1) we can substitute $h^{2}+x^{2}=p^{2}$ in equation (8),

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

$c^{2}=p^{2}-a x+\frac{a^{2}}{4}$....(9)

(iii) Now we will add equations $(5)$ and $(9)$ as shown below,

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

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

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

Therefore, $b^{2}+c^{2}-2 p^{2}+\frac{a^{2}}{2}$.