in the given figure abcd is a square and angle pqr=90° if PB=QC=DR prove that qb=rc and pq=qr and angle qpr = 45°
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Since PBCR becomes rectangle
∴ PB=RC=x
⇒AP=x
If QC=RC=x ∴∠CRQ=∠RCQ=45^{0}450
Since ∠PRC=90^{0}900
∴∠PRQ=45^{0}450
and sum of angles a triangle is 180^{0}1800
∴∠PQR=90^{0}900 and ∠PRQ=45^{0}450
∴∠QPR=180^{0}1800 -(90^{0}900 +45^{0}450 )=45^{0}450
HOPE IT HELPS !
∴ PB=RC=x
⇒AP=x
If QC=RC=x ∴∠CRQ=∠RCQ=45^{0}450
Since ∠PRC=90^{0}900
∴∠PRQ=45^{0}450
and sum of angles a triangle is 180^{0}1800
∴∠PQR=90^{0}900 and ∠PRQ=45^{0}450
∴∠QPR=180^{0}1800 -(90^{0}900 +45^{0}450 )=45^{0}450
HOPE IT HELPS !
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