(a) AB2 + AC2 = 2AD2 + 2BD2
Let AD be the median of ΔABC
∴ BD = DC
Let AE be the perpendicular from A on BC. Then,
In rt. ΔABE,
AB2 = BE2 + AE2
= (BD – ED)2 + AE2
= BD2 + AD2 – 2BD.ED …(i)
In rt. ΔACE,
AC2 = AE2 + EC2 = AE2 + (ED + DC)2
= AD2 + DC2 + 2ED.DC
= AD2 + BD2 + 2ED.BD (∵ BD = DC) …(ii)
Adding (i) and (ii), we get
AB2 + AC2 = 2AD2 + 2BD2.