In the given figure, abcd is …

Given: ABCD is a square, where $\angle$PQR = 90^o and PB = QC = DR
To prove: (i) QB = RC (ii) PQ = QR
(iii) $\angle$QPR = 45^o
Proof:

1. Here,
   BC = CD … (Sides of square)
   CQ = DR … (Given)
   BC = BQ + CQ
   $\therefore$ CQ = BC − BQ
   $\therefore$ DR = BC – BQ .......(1)
   Also,
   CD = RC + DR
   $\therefore$ DR = CD - RC = BC - RC ........(2)
   From (1) and (2), we have,
   BC - BQ = BC - RC
   $\therefore$ BQ = RC
2. Now in $\triangle$RCQ and $\triangle$QBP
   we have, PB = QC … (Given)
   BQ = RC … [from (i)]
   $\angle$RCQ = $\angle$QBP … 90^o each
   Hence by SAS(Side-Angle-Side) congruence rule,
   $\triangle$RCQ $\cong$ $\triangle$QBP
   $\therefore$ QR = PQ … (by cpct)
3. $\triangle$RCQ $\cong$ $\triangle$QBP and QR = PQ … [from (2)]
   $\therefore$ In $\triangle$RPQ,
   $\angle$QPR = $\angle$QRP =$\frac{1}{2}$(180^o - 90^o) = $\frac{90}{2}$= 45^o
   $\therefore$ $\angle$QPR = 45^o