#### Question

In ∆ABC, seg AD ⊥ seg BC DB = 3CD. Prove that :

2AB^{2 }= 2AC^{2 }+ BC^{2}

^{}

#### Solution

It is given that,

DB = 3 CD

∴ BC = 4 CD ....(1)

According to Pythagoras theorem,

In ∆ABD

\[ \Rightarrow {AD}^2 = {AB}^2 - {DB}^2 . . . \left( 2 \right)\]

In ∆ACD

\[{AC}^2 = {AD}^2 + {CD}^2 \]

\[ \Rightarrow {AC}^2 = \left( {AB}^2 - {DB}^2 \right) + {CD}^2 \left( \text{From} \left( 2 \right) \right) \]

\[ \Rightarrow {AC}^2 = {AB}^2 - \left( 3CD \right)^2 + {CD}^2 \left( \text{Given} \right)\]

\[ \Rightarrow {AC}^2 = {AB}^2 - 9 {CD}^2 + {CD}^2 \]

\[ \Rightarrow {AC}^2 = {AB}^2 - 8 {CD}^2 \]

\[ \Rightarrow {AB}^2 = {AC}^2 + 8 {CD}^2 \]

\[ \Rightarrow {AB}^2 = {AC}^2 + 8 \left( \frac{BC}{4} \right)^2 \left( \text{From} \left( 1 \right) \right)\]

\[ \Rightarrow {AB}^2 = {AC}^2 + \frac{{BC}^2}{2}\]

\[ \Rightarrow 2 {AB}^2 = 2 {AC}^2 + {BC}^2\]

Hence, 2AB^{2 }= 2AC^{2 }+ BC^{2}.