In the picture, chords AB and CD of the circle are extended to meet at P.

i) Prove that the angles of □ APC and □ PBD, formed by joining AC and BD, are the same.

ii) Prove that PA × PB = PC × PD.

iii) Prove that if PB = PD, then ABCD is an isosceles trapezium.

(i)

As, ABDC is a cyclic quadrilateral

⇒ ∠ACD + ∠ABD = 180° [Sum of opposite pair of angles in a cyclic quadrilateral is 180°]

⇒ ∠ABD = 180° – ∠ACD

Also,

∠ABD + ∠PBD = 180° [linear pair]

⇒180 – ∠ACD + ∠PBD = 180°

⇒∠ACD = ∠PBD …[1]

Similarly

⇒∠BAC = ∠PDB …[2]

And, clearly

∠BPD = ∠APC [Common] …[3]

From [1], [2] and [3]

It's clear that all angles of both the triangles are equal.

And ΔAPC ~ ΔBPD [By AAA similarity criterion]

(ii) As,

ΔAPC ~ ΔBPD

⇒ PD × PC = PB × PA

Or

⇒ PA × PB = PC × PD

(iii) As, PD = PB …[4]

∠PBD = ∠PDB [Angles opposite to equal sides are equal] …[5]

and also, as ΔAPC ~ ΔBPD

⇒ ∠ACD = ∠PBD

⇒ ∠ACD = ∠PDB [From 5]

⇒ AC || BD [As corresponding angles are equal]

Also,

⇒ PD × PC = PB × PA [from ii part]

⇒ PD × PC = PD × PA

⇒ PC = PA …[6]

On subtracting [6] from [5]

⇒ PC – PD = PA – PB

⇒ CD = AB

Hence, in quadrilateral, ABDC one pair of opposite sides is parallel, and sides of another pair are equal,

Therefore, ABDC is an isosceles trapezium.