Given: ABCD is a square and PB=PD

To prove: CPA is a straight line

Proof:

∆APD and ∆APB,

DA = AB …as ABCD is square

AP = AP … common side

PB = PD … given

Thus by SSS property of congruence,

∆APD ≅ ∆APB

Hence, we know that, corresponding parts of the congruent triangles are equal

∠APD = ∠APB …(1)

Now consider ∆CPD and ∆CPB,

CD = CB … ABCD is square

CP = CP … common side

PB = PD … given

Thus by SSS property of congruence,

∆CPD ≅ ∆CPB

Hence, we know that, corresponding parts of the congruent triangles are equal

∠CPD = ∠CPB … (2)

Now,

Adding both sides of 1 and 2,

∠CPD +∠APD = ∠APB + ∠CPB …(3)

Angels around the point P add upto 360°

∴ ∠CPD +∠APD + ∠APB + ∠CPB = 360°

From 4,

2(∠CPD +∠APD) = 360°

∠CPD +∠APD = = 180°

This proves that CPA is a straight line.