In the given figure, ABCD is a square and P is a point inside it such that PB=PD. Prove that CPA is a straight line.
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.