In the figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that (i) \(\triangle\) PAC ~ \(\triangle\) PDB (ii) PA.PB = PC.PD

Solution :

(i)  In \(\triangle\)s PAC and PDB, we have :

\(\angle\) APC = \(\angle\) DPB         (common)

\(\angle\) PAC = \(\angle\) PDB

[\(\therefore\)  \(\angle\) BAC = 180 – \(\angle\) PAC  and \(\angle\) PDB = \(\angle\) CDB = 180 – (180 – \(\angle\) PAC) = \(\angle\) PAC]

\(\therefore\)  By AA similarity, we have :

\(\triangle\) PAC ~ \(\triangle\) PDB

(ii)  Since \(\triangle\)s PAC ~ PDB, therefore

\(PA\over PD\) = \(PC\over PB\)  or  PA.PB = PC.PD.