Question

P is any point inside the rectangle ABCD. Prove that PA² + PC² = PB² + PD².

Answer 1

Given: Rectangle ABCD and a point P inside the rectangle.

To prove: PA² + PC² = PB² + PD²

Proof: Let us draw a line PQ through O, which is parallel to BC.

Hence OQ || AD

⇒ OQ || AD

All angles of a rectangle are 90º.

Since OQ || BC

AB is the transversal.

∠AOP = ∠B

∠AOP = 90º

Similarly,

We can prove

∠BOP = 90º, ∠DQP = 90º and ∠CQP = 90º

Using Pythagoras theorem,

Hypotenuse² = Height² + Base²

In right angled triangle ΔOPB,

PB² = BO² + PO²   ...(1)

In right triangle ΔPQD,

PD² = PQ² + DQ²  ...(2)

Similarly,

In right triangle ΔPQC,

PC² = PQ² + CQ²   ...(3)

In right triangle ΔPAD,

PA² = AD² + PD²   ...(4)

Adding equation (1) and (2),

PB² + PD² = BO² + PO² + PQ² + DQ²   ...(As PQ || BC)

BQ || CQ  ...(Because AB || CD)

BOQC is a parallelogram.

So, BO = AO   ...(As opposite side of a parallelogram are equal.)

= CQ² + PO² + PQ² + AO²

= CQ² + PQ² + PO² + AO²

= PC² + PA²

Thus, PB² + PD² = PC² + PA²

Or

PC² + PA² = PB² + PD²

Hence proved.