Question:

In right $\Delta ABD$: $AB^2 = AD^2 + BD^2$ -- (1). [1.0 Mark]
In right $\Delta ACD$: $AC^2 = AD^2 + CD^2$ -- (2). [1.0 Mark]
Subtracting (2) from (1): $AB^2 - AC^2 = BD^2 - CD^2$. [1.0 Mark]
Given $BD = 3 CD$. Also $BC = BD + CD = 3 CD + CD = 4 CD \Rightarrow CD = \dfrac{BC}{4}$ and $BD = \dfrac{3 BC}{4}$. [1.0 Mark]
$AB^2 - AC^2 = \left(\dfrac{3 BC}{4}\right)^2 - \left(\dfrac{BC}{4}\right)^2 = \dfrac{9 BC^2}{16} - \dfrac{BC^2}{16} = \dfrac{8 BC^2}{16} = \dfrac{BC^2}{2}$.
$2(AB^2 - AC^2) = BC^2 \Rightarrow 2 AB^2 - 2 AC^2 = BC^2 \Rightarrow 2 AB^2 = 2 AC^2 + BC^2$. Proved! [1.0 Mark]

---
🎯 Official CBSE Marking Scheme:
Pythagoras in $\Delta ABD$ ($AB^2 = AD^2 + BD^2$): 1.0 Mark
Pythagoras in $\Delta ACD$ ($AC^2 = AD^2 + CD^2$): 1.0 Mark
Subtracting equations to eliminate $AD^2$: 1.0 Mark
Expressing $BD$ and $CD$ in terms of $BC$: 1.0 Mark
Simplifying to $2 AB^2 = 2 AC^2 + BC^2$: 1.0 Mark