Question

Two chords AB and CD of a circle intersect at a point P outside the circle. Prove that (a) ΔPAC ~ ΔPDB (b) PA · PB = PC · PD. 

Answer 1

Given: AB and CD are two chords
To Prove:
(a) Δ PAC - Δ PDB
(b) PA. PB = PC.PD
Proof: ∠𝐴𝐵𝐷 + ∠𝐴𝐶𝐷 = 180° …(1) (Opposite angles of a cyclic quadrilateral are
supplementary)
∠𝑃𝐶𝐴 + ∠𝐴𝐶𝐷 = 180° …(2)                     (Linear Pair Angles )
Using (1) and (2), we get
∠𝐴𝐵𝐷 = ∠𝑃𝐶𝐴
∠𝐴 = ∠𝐴                                                          (Common)  

By AA similarity-criterion Δ PAC - Δ PDB
When two triangles are similar, then the rations of the lengths of their corresponding sides are proportional  

`∴ (PA)/(PD)=(PC)/(PB)` 

⟹ PA.PB = PC.PD