O is any point inside a rectangle ABCD.
Prove that OB2 + OD2 = OA2 + OC2
Sol:
PQ ║ BC
PQ ⊥ AB
and
PQ ⊥ DC
(∠B = 90° and ∠C = 90°)
∠BPQ
=
90º
∠CQP
=
90º
and
∴ BPQC and APQD are both rectangles.
In ΔOPB,
OB2
=
BP2
+
OP2
(1)
In ΔOQD,
OD2
=
OQ2
+
DQ2
(2)
In ΔOQC,
OC2
=
OQ2
+
CQ2
(3)
In ΔOAP,
OA2
=
AP2
+
OP2
(4)
Ex.6.5(Q.8)
A
P
B
C
D
Q
O
Construction : through Q, draw PQ ║ BC intersecting
AB and CD at P and Q respectively.
∴
[by Pythagoras theorem]
}
OB2 + OD2 = OA2 + OC2
OB2 + OD2 = OA2 + OC2
O is any point inside a rectangle ABCD.
Prove that OB2 + OD2 = OA2 + OC2
Sol:
Adding (1) and (2),
OB2
=
BP2
+
OP2
OD2
+
+
OQ2
+
DQ2
=
CQ2
+
OP2
+
OQ2
+
AP2
(As BP = CQ and DQ = AP)
=
CQ2
+
OQ2
+
OP2
+
AP2
=
OC2
+
OA2
[From (3) and (4)]
A
P
B
C
D
Q
O
OB2 = BP2 + OP2 …(1)
OD2 = OQ2 + DQ2 …(2)
OC2 = OQ2 + CQ2 …(3)
OA2 = AP2 + OP2 …(4)
OB2
OD2
+