Solution: In ∆ ACE; AE2 = AC2 + CE2 ……… (1)
In ∆ DCB; BD2 = DC2 + CB2 ……… (2)
In ∆ ACB; AB2 = AC2 + CB2 ……… (3)
In ∆ DCE; DE2 = DC2 + CE2 ……… (4)
Adding equations (1) and (2), we get;
AE2 + BD2 = AC2 + CE2 + DC2 + CB2 …….. (5)
Adding equations (3) and (4), we get;
AB2 + DE2 = AC2 + CB2 + DC2 + CE2 ………. (6)
On comparing the RHS of equations (5) and(6), we get;
AE2 + BD2 = AB2 + DE2 proved