In the given Fig, PA, QB and RC are each perpendicular to AC. Prove that 

In ∆CBQ and ∆CAP
∠BCQ = ∠ACP [common]
and    ∠CBQ = ∠CAP = 90°
Therefore, by AA similar condition
                              
                               

In ∆ABQ and ∆ACR
∠BAQ = ∠CAR [common]
and    ∠ABQ = ∠ACR = 90°
Therefore, by AA similar condition
           
                                 ...(ii)
Adding (i) and (ii), we get