Consider  $\mathrm{△}ADB$ and  $\mathrm{△}ABC$

∠BAD = ∠BAC                [common angle]

∠BDA = ∠ABC                [ ${90}^{∘}$]

Therefore by AA similarity criterion, $\mathrm{△}ADB$ and  $\mathrm{△}ABC$ are similar.

So,  $\frac{AB}{AC}=\frac{AD}{AB}$  ⇒  ${AB}^2$ = AC.AD   ---(I)

Similarly, $\mathrm{△}BDC$ and  $\mathrm{△}ABC$ are similar.

So, $\frac{BC}{AC}$ =  $\frac{DC}{BC}$  ⇒  ${BC}^2$ = AC.DC   ---(II)

Dividing (I) and (II) and cancelling out AC, we get,   ${\left(\frac{AB}{BC}\right)}^2$ =   $\frac{AD}{DC}$  ----(III)

Also, In $\mathrm{△}ADB$, ${AB}^2$ = ${AD}^2$ + ${DB}^2$ and in  $\mathrm{△}BDC$, ${CB}^2$ = ${CD}^2$ + ${DB}^2$ [Pythagoras theorem]

Subtracting these two equations above and cancelling off ${DB}^2$ on both sides, we get 

${AB}^2$ - ${BC}^2$ = ${AD}^2$ - ${CD}^2$ ⇒    ${AB}^2$ + ${CD}^2$ = ${AD}^2$  + ${BC}^2$  --------------(IV)

Dividing this equation with ${BC}^2$ on both sides,  $\frac{{AB}^2}{{BC}^2}$ +  $\frac{{CD}^2}{{BC}^2}$ =  $\frac{{AD}^2}{{BC}^2}$ +  $\frac{{BC}^2}{{BC}^2}$   ⇒  $\frac{{AB}^2}{{BC}^2}$ + $\frac{{CD}^2}{{BC}^2}$ = $\frac{{AD}^2}{{BC}^2}$ + 1

$⟹$  $\frac{{AB}^2}{{BC}^2}$ - 1 = $\frac{{AD}^2}{{BC}^2}$ - $\frac{{CD}^2}{{BC}^2}$

$⟹$ $\frac{AD}{DC}$ - 1 =  $\frac{{AD}^2−{CD}^2}{{BC}^2}$    [Using (III)]

$⟹$$\frac{AD−DC}{DC}$ = $\frac{(AD−CD)(AD+CD)}{{BC}^2}$

$⟹$$\frac{1}{DC}$ = $\frac{AC}{{BC}^2}$ 

$⟹$ ${BC}^2$ = AC.DC

Alternatively,

Consider $\mathrm{△}ABC$ and $\mathrm{△}BDC$,

$\mathrm{âˆ}ABC=\mathrm{âˆ}BDC={90}^{∘}$

$\mathrm{âˆ}C=\mathrm{âˆ}C$    [Common angle]

Therefore by AA similarity criterion, $\mathrm{△}ABC$ and  $\mathrm{△}BDC$ are similar.

$\frac{AC}{BC}$ = $\frac{BC}{DC}$

$B{C}^2$ = $AC×DC$