In the given Fig., AD ⊥ BC. Prove that AB2 + CD2 = BD2 + AC2.
In right ∆ADC. by Pythagoras Theorem,
AC2 = AD2 + CD2 ...(i)
In right ∆ADB, by Pythagoras Theorem,
AB2 = AD2 + BD2 ...(ii)
Subtracting (i) from (ii), we get
AB2 - AC2 = BD2 - CD2
⇒ AB2 + CD2 = BD2 + AC2 Hence Proved.