If in $Δ A B C$ , $A D$ is median and $A M ⊥ B C$ , then prove that ${A B}^{2} + {A C}^{2} = 2 {A D}^{2} + \frac{1}{2} {B C}^{2}$ .

Open in App

Solution

In $△ A D B,$
${A B}^{2} = {A D}^{2} + {B D}^{2}$ --(1)

In $△ A D C,$
${A C}^{2} = {A D}^{2} + {C D}^{2}$ --(2)

Adding (1) and (2),
${A B}^{2} + {A C}^{2} = {2 A D}^{2} + {B D}^{2} + {C D}^{2} {A B}^{2} + {A C}^{2} = {2 A D}^{2} + 2 {C D}^{2}$

But BC = 2CD
So, ${A B}^{2} + {A C}^{2} = {2 A D}^{2} + 2 {(1 / 2 B C)}^{2} {A B}^{2} + {A C}^{2} = {2 A D}^{2} + 2 (1 / 4 {B C}^{2}) {A B}^{2} + {A C}^{2} = {2 A D}^{2} + 1 / 2 {B C}^{2}$

Hence proved.

Suggest Corrections

Similar questions

△ A B C, A D

A E ⊥ B C

A B^{2} + A C^{2} = 2 A D^{2} + \frac{1}{2} B C^{2}

A D

A B C

A M ⊥ B C

A C^{2} = A D^{2} + B C . D M + {(\frac{B C}{2})}^{2}

A B^{2} = A D^{2} B C . D M + {(\frac{B C}{2})}^{2}

A C^{2} + A B^{2} = 2 A D^{2} + \frac{1}{2} B C^{2}

Δ A B C,

A B^{2} + A C^{2} = 2 A D^{2} + 2 B D^{2} ?

{AB}^{2} + {AC}^{2} = 2 ({AD}^{2} + {CD}^{2})

Join BYJU'S Learning Program