In the given figure, abcd is …

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

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