In the figure, the sides AB and BC of square ABCD are produced to P and Q respectively so that BP = CQ. Prove that DP and AQ are perpendicular to each other.

Since ABCD is a square,  AB = BC
    Also BP = CQ ...... (given)
.^.. AB + BP = BC + CQ
       .^.. AP = BQ 
Now in ΔAPD and ΔBQA,  
          AP = BQ ...... (proved above)
      ∠ABQ = ∠DAP ... (each 90°)
    and AB = AD
   .^.. ΔAPD ≅ ΔBQA
  .^.. ∠APD = ∠BQA ...... (c.p.c.t.)
and ∠ADP = ∠QAP ...... (c.p.c.t.)
Also ∠DAQ = ∠AQB
  .^.. ∠DAO = ∠APO
Now in ΔAOD and ΔAOP,  
     ∠ADO = ∠OAP,
and ∠DAO = ∠APO
       .^.. 3rd ∠DOA = 3rd ∠AOP  [Since, two angles of DAOD and  DAOP are equal, the third angle is also equal]
but ∠DOA + ∠AOP = 180°
          .^.. 2 ∠DOA = 180°
             .^.. ∠DOA = 90°
.^.. DO is perpendicular to AO or DP is perpendicular to AQ.