Question
In a circle, two chords AB and CD intersect a point P inside the circle. Prove that
  1. $\triangle\text{PAC}\sim\triangle\text{PDB}$
  2. $\text{PA}.\text{PB}=\text{PC}.\text{PD}$

Answer

In $\triangle\text{PAC}$ and $\triangle\text{BPD},$
$\angle\text{APC}=\angle\text{BPD}$ .....(Vertically opposite angles)
$\angle\text{CAP}=\angle\text{BDP}$ ......(Angels inscribed in the same are equal)
$\therefore\triangle\text{PAC}\sim\triangle\text{PDB}$ .....(AA similarity criterion)
$\Rightarrow\frac{\text{PA}}{\text{PD}}=\frac{\text{PC}}{\text{PB}}$
$\Rightarrow\text{PA}.\text{PB}=\text{PC}.\text{PD}$

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