Question

In $\Delta ABC,$ AD is the median prove that $A{B}^2+A{C}^2=2A{D}^2+2B{D}^2?$

Answer 1

Given : $△$ ABC with a median AD.

To prove : $A{B}^2+A{C}^2=2A{D}^2+2B{D}^2$

Construction : Draw AN perpendicular to BC to meet BC at N.

Proof :

Case 1 :

If $\angle ADB=\angle ADC$ then they are right angles.

Here,

$A{B}^2=B{D}^2+A{D}^2$

$A{C}^2=C{D}^2+A{D}^2$

Adding these equations, we get,

$A{B}^2+A{C}^2=2A{D}^2+2B{D}^2$ [Since BD=CD]

Case 2 :

When $\angle ADB\ne \angle ADC$

Since, $\angle ADB$ is obtuse,

$A{B}^2=B{D}^2+A{D}^2+2BD.DN$ ...(1)

Since, $\angle ADC$ is acute,

$A{C}^2=D{C}^2+A{D}^2-2DC.DN$ ....(2)

But $BD=DC$

Adding 1 & 2 , we get,

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