Q29 of 40 Page 1

In a rectangle ABCD, P is any interior point. Then prove that PA2 + PC2 = PB2 + PD2.


Draw EF || AB || DC passing through P


Also, AB = EF = DC


∴ ABCD and CDEF are rectangles


Now by Pythagoras theorem,


PB2 = BF2 + PF2 …(1)


and,


PD2 = PE2 + ED2 …(2)


Adding (1) and (2),


PB2 + PD2 = BF2 + PF2 + PE2 + ED2


⇒ PB2 + PD2 = AE2 + PF2 + PE2 + CF2


⇒ PB2 + PD2 = AE2 + PE2 + PF2 + CF2


⇒ PB2 + PD2 = PA2 + PC2


Hence, Proved


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