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)