The perpendicular from $A$ on side $BC$ of a $\Delta ABC$ intersects $BC$ at $D$ such that $DB = 3CD$ (see Figure). Prove that $2AB^{2} = 2AC^{2} + BC^{2}$.

(N/A) Applying the Pythagoras theorem for $\triangle ACD$,we obtain:
$AC^{2} = AD^{2} + DC^{2}$
$AD^{2} = AC^{2} - DC^{2}$ $...(1)$
Applying the Pythagoras theorem for $\triangle ABD$,we obtain:
$AB^{2} = AD^{2} + DB^{2}$
$AD^{2} = AB^{2} - DB^{2}$ $...(2)$
From equation $(1)$ and equation $(2)$,we obtain:
$AC^{2} - DC^{2} = AB^{2} - DB^{2}$ $...(3)$
It is given that $DB = 3CD$. Since $BC = DB + CD$,we have $BC = 3CD + CD = 4CD$.
$\therefore CD = \frac{BC}{4}$ and $DB = 3CD = \frac{3BC}{4}$.
Substituting these values into equation $(3)$,we obtain:
$AC^{2} - \left(\frac{BC}{4}\right)^{2} = AB^{2} - \left(\frac{3BC}{4}\right)^{2}$
$AC^{2} - \frac{BC^{2}}{16} = AB^{2} - \frac{9BC^{2}}{16}$
$16AC^{2} - BC^{2} = 16AB^{2} - 9BC^{2}$
$16AB^{2} - 16AC^{2} = 8BC^{2}$
Dividing by $8$,we get:
$2AB^{2} = 2AC^{2} + BC^{2}$