Thank you vansh. my question is in the given figure ABCD is a square and angle pqr is equal to 90 degree if pb is equal to Qc is equal to DR, prove that QB is equal to RC , PQ is equal to QR , angle QPR is equal to 45 degree .
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Answer:
ABCD is a square
B
C
=
C
D
G
i
v
e
n
Q
C
=
R
D
∴
B
C
−
Q
C
=
C
D
−
R
D
⇒
B
Q
=
C
R
(ii)
In
Δ
P
B
Q
and \Delta QCR\)
P
B
=
Q
C
[given]
B
Q
=
C
R
[Proved in (a)]
∠
P
B
Q
=
∠
Q
C
R
[Each is
90
∘
]
∴
Δ
P
B
Q
and
Δ
Q
C
R
[SAS Congruency]
⇒
P
Q
=
Q
R
,
∠
B
P
Q
=
∠
C
Q
R
,
∠
B
Q
P
=
∠
C
R
Q
[C.P.C.T.]
BQC is a straight line
∴
∠
B
Q
P
+
∠
P
Q
R
+
∠
C
Q
R
=
180
∘
⇒
∠
B
Q
P
+
∠
P
Q
R
+
∠
B
P
Q
=
180
∘
⇒
(
∠
B
Q
P
+
∠
B
P
Q
)
+
∠
P
Q
R
=
180
∘
⇒
180
∘
−
∠
P
B
Q
+
∠
P
Q
R
=
180
∘
[Angle sum property for
Δ
P
B
Q
]
⇒
180
∘
−
90
∘
+
∠
P
Q
R
=
180
∘
⇒
∠
P
Q
R
=
90
∘
(iii)
Δ
P
Q
R
is an isosceles right angled triangle, as
P
Q
=
Q
R
and
∠
P
Q
R
=
90
∘
.
⇒
∠
P
Q
R
+
∠
Q
R
P
+
∠
Q
P
R
=
180
∘
⇒
90
∘
+
2
∠
Q
P
R
=
180
∘
∴
∠
Q
P
R
=
45
∘
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