Question

in figure ABCD is a square. if angle PQR=90 and PB=QC=DR, prove that angle QPR=45.
Plz answer it fast i need it right now

Answer 1

Since PBCR becomes rectangle 
∴ PB=RC=x 
⇒AP=x
If QC=RC=x ∴∠CRQ=∠RCQ=$45^{0}$
Since ∠PRC=$90^{0}$
∴∠PRQ=$45^{0}$
and sum of angles a triangle is $180^{0}$
∴∠PQR=$90^{0}$  and ∠PRQ=$45^{0}$
∴∠QPR=$180^{0}$-($90^{0}$+$45^{0}$)=$45^{0}$

Answer 2

we can say thatCD = BC(sides of square).....eq.1

CD=CR + DR......eq2

BC= BQ + QC......eq3

from all the above 3 equations we can easily said that CR+DR=BQ+QC

but DR=QC (given)

so..CR=BQ

Now in triangle PBQ and triangle RCQ

PB=QC (given)

angle PBQ= angle RCQ (each of 90°)

BQ=RC (proved above)

by SAS congruence axiom triangle PBQ and triangle RCQ are congruent

this implies that PQ=RQ(cpct)

in triangle PQR

PQ=RQ(proved above)

so angle QRP = angle QPR as angles opp. to equal sides are equal

by angle sum property of triangle PQR

angle PQR + angle QRP + angle QPR = 180°

90° + 2 (angle QPR) = 180° ( it is given that angle PQR =90° nd it is proved above that angle QPR = angle QRP)

2(angle QPR)= 180° - 90°= 90°

Angle QPR=90°÷ 2 =45°

HENCE PROVED

hope that it helps u....