Question

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

Answer 1

Given: P is any point in the interior of rectangle ABCD.

To Prove: PA² + PC² = PB² + PD².

Construction: Draw a line parallel to AB and CD.

Proof: AB || MN and AM || BN, Also, A = 90°

∴ ABNM is rectangle.

Also, MNCD is a rectangle.

Here, PM ⊥^r AD and PN ⊥^r BC

And AM = BN and MD = NC   ...(i)

Now, in ΔAMP, AP² = AM² + MP²   ...(ii)

In ΔPMD, PD² = MP² + MD²   ...(iii)

In ΔPNB, PB² = PN² + BN²   ...(iv)

In ΔPNC, PC² = PN² + NC²   ...(v)

∴ PA² + PC² = AM² + MP² + PN² + NC²   ...[From equations (ii) and (v)]

= BN² + MP² + PN² + MD²   ...[From equation (i)]

= (BN² + PN²) + (MP² + MD²)

= PB² + PD²   ...[From equations (iii) and (iv)]

Hence Proved.