for any triangle ABC,prove that AB^2+AC^2=2(AD^2+BD^2) given that D is the midpoint of bc.

Question

the question is from plane geometry

Answer 1

Question: for any triangle ABC,prove that AB^2+AC^2=2(AD^2+BD^2) given that D is the midpoint of bc.

We are given that,

DC = BD   ---------------------------------- (1)

angle ADC = pi - angle ADB

Hence cos(ADC) = cos(pi - ADB) = cos(pi).cos(ADB) + sin(pi).sin(ADB)

cos(ADC) = -cos(ADB)  ---------------- (2)

Using the cosine rule, twice

AB^2 = AD^2 + BD^2 - 2.AD.BD.cos(ADB)   --------------------- (3)
AC^2 = AD^2 + DC^2 - 2.AD.DC.cos(ADC)  --------------------- (4)

Substituting for (1) and (2) into (4),

AB^2 = AD^2 + BD^2 - 2.AD.BD.cos(ADB)   --------------------- (5)
AC^2 = AD^2 + BD^2 + 2.AD.DC.cos(ADB)  --------------------- (6)

Adding (5) and (6),

AB^2 + AC^2 = 2(AD^2 + BD^2)