Question

In a circle, two chords AB and CD intersect at a point P inside the circle. Prove that
(a) ΔPAC ∼PDB (b) PA. PB= PC.PD  

 

Answer 1

Given : AB and CD are two chords
To Prove:
(a) Δ PAC ~ ΔPDB
(b) PA.PB = PC.PD
Proof: In Δ PAC and Δ PDB
∠𝐴𝑃𝐶 = ∠𝐷𝑃𝐵 (𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑙𝑦 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑛𝑔𝑙𝑒𝑠)
∠𝐶𝐴𝑃 = ∠𝐵𝐷𝑃 (𝐴𝑛𝑔𝑙𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙)
𝑏𝑦 𝐴𝐴 𝑠𝑖𝑚𝑖𝑙𝑎𝑟𝑖𝑡𝑦 𝑐𝑟𝑖𝑡𝑒𝑟𝑖𝑜𝑛 Δ𝑃𝐴𝐶 ~𝑃𝐷𝐵
When two triangles are similar, then the ratios of lengths of their corresponding sides are proportional.  

`∴ (PA)/(PD)=(PC)/(PB)` 

⟹ PA. PB = PC. PD