In the given figure, ABCD is a square and `angle PQR=90^(@)." If " PB=QC=DR,` prove that
(i) `QB=RC, " (ii) " PQ =QR, " (iii) " angle QPR=45^(@).`

Explanation

`BC=DC, CQ=DR rArr BC-CQ=DC-DR rArr QB=RC.`
From `triangle CQR, angle RQB= angle QCR +angle QRC`
`rArr angle RQP+angle PQB=90^(@) + angle QRC`
`rArr 90^(@) + angle PQB=90^(@)+angle QRC rArr angle PQB= angle QRC.`
Now, `triangle RCQ~= triangle QBP " and therefore , "QR=PQ. `
`PQ =QR rArr angle QPR = angle PRQ.`
But, `angle QPR+angle PRQ=90^(@)." So, " angle QPR=45^(@).`