A line drawn from vertex A of an equilateral triangle ABC, meets BC at D and circumcircle of triangle ABC at P. If BP = 5, PC = 20 then PD = ________

  AC – subtends ${60}^0$  at B
$\Rightarrow$ AC – subtends  ${60}^0$at P
Let  $\overline{)PAC}=\theta \Rightarrow \overline{)PBC}=\theta$
 $\overline{)BCA}={60}^0\Rightarrow \overline{)BPA}={60}^0$

 $\therefore {\Delta }^{le}BPD,{\Delta }^{le}APC$ are similar                  
 $\frac{PA}{PB}=\frac{PC}{PD}$
 $PA\times PD=PB\times PC\to \left(1\right)$
From (2) $PA=PB+PC$                        By ptolmy’s theorem
$\therefore PB\times PC=(PB+PC)PD$                Sine ABPC is cyclic quatrilateral
   $PB\times PC=PB\times PD+PC\times PD$           $PA\times BC=PBAC+PC.AB$
     $\frac{1}{PD}=\frac{1}{PC}+\frac{1}{PB}$                                       $PA\times BC=(PB+PC)AC$
         $\frac{1}{PD}=\frac{1}{20}+\frac{1}{5}=\frac{4+1}{20}=\frac{5}{20}=\frac{1}{4}$                 
                                 $BC=AC(\therefore {\Delta }^{le}ABCisequilateral)$
$PD=4$                           $PA=PB+PC\to \left(2\right)$

IInd method. Apply sine – rule in ${\Delta }^{le}BPD,$  and  ${\Delta }^{le}PDC,$
$\frac{5}{\sin (120-\theta )}=\frac{PD}{\sin \theta }\Rightarrow PD=\frac{5\sin \theta }{\sin (60+\theta )}$  and  $\frac{20}{\sin (60+\theta )}=\frac{PD}{\sin (60-\theta )}$
 $PD=\frac{20\sin (60-\theta )}{\sin (60+\theta )}$
 $\Rightarrow \frac{\sin (60-\theta )}{\sin \theta }=\frac{1}{4}\Rightarrow \cos \theta =\frac{\sqrt{3}}{2}$
 $\Rightarrow PD=\frac{5\sin \theta }{\frac{\sqrt{3}}{2}\sin \theta +\frac{1}{2}\sin \theta }=4$