$△ A B C$ is right angled at $B$ and $D$ is the midpoint of $B C$ .Prove that $A C^{2} = 4 A D^{2} - 3 A B^{2}$

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Solution

In $△ A B C$ , using the Pythagoras theorem, we get

${A C}^{2} = {A B}^{2} + {B C}^{2} = {A B}^{2} + {(2 B D)}^{2} = {A B}^{2} + 4 {B D}^{2}$

$\frac{{A C}^{2} - {A B}^{2}}{4} = {B D}^{2}$

In $△ A B D$

${A D}^{2} = {A B}^{2} + {B D}^{2}$ (Pythagoras theorem)
So, ${A D}^{2} = {A B}^{2} + \frac{{A C}^{2} - {A B}^{2}}{4}$

So, $4 {A D}^{2} = 3 {A B}^{2} + {A C}^{2}$

So, ${A C}^{2} = 4 {A D}^{2} - 3 {A B}^{2}$

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