i. Prove that the angles of Δ APC and Δ PBD, formed by joining AC and BD, are the same. ii. Prove that PA × PB = PC × PD.

Question

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

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

ii. Prove that PA × PB = PC × PD. 

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

Answer 1

i. As ABCD is a cyclic quadrilateral. 

If ∠BAC = x° then ∠BDC = 180 – x ∠BDP = x°

If ∠ACD = y° then ∠PBD = y° As ∠APC is common angle. 

Angles of Δ APC and Δ PBD are same

ii. As △APC, △PBD are similar triangles,

\(\frac{AP}{PD}=\frac{PC}{PB}\) AP x PB = PC x PD

iii. If PB = PD ; AP = PC 

In ABCD

ABDC is a cyclic quadrilateral, so their opposite angles are supplementary. 

If AP = PC, in ∆ APC 

∠A =∠C

As AB = CD 

AC || BD 

Adjacent angles are supplementary 

∴ ABCD will be an isosceles trapezium