the diagonal AC and BD of a rohombus intersect each other at O prove that AB square +BC square +CD square +DA square =4( OA square +OB square)
Answers
Step-by-step explanation:
Since, diagonals of a Rhombus intersect each other at right angels.
∴ in ΔAOB,
AB
2
=OA
2
+OB
2
........... (1)
and in ΔBOC,
BC
2
=OB
2
+OC
2
........... (2)
and in ΔCOD,
CD
2
=OC
2
+OD
2
........... (3)
and in ΔAOD,
AD
2
=OA
2
+OD
2
........... (4)
Above results are obtained by using Pythagorous theorem.
Now, (1)+(2)+(3)+(4) gives
AB
2
+BC
2
+CD
2
+AD
2
=2(OA
2
+OB
2
+OC
2
+OD
2
)
=4(OA
2
+OB
2
) ...... [Since, OD=OB & OA=OC]
∴AB
2
+BC
2
+CD
2
+AD
2
=4(OA
2
+OB
2
)
solution
Step-by-step explanation:
Since, diagonals of a Rhombus intersect each other at right angels.
∴ in ΔAOB,
AB
2
=OA
2
+OB
2
........... (1)
and in ΔBOC,
BC
2
=OB
2
+OC
2
........... (2)
and in ΔCOD,
CD
2
=OC
2
+OD
2
........... (3)
and in ΔAOD,
AD
2
=OA
2
+OD
2
........... (4)
Above results are obtained by using Pythagorous theorem.
Now, (1)+(2)+(3)+(4) gives
AB
2
+BC
2
+CD
2
+AD
2
=2(OA
2
+OB
2
+OC
2
+OD
2
)
=4(OA
2
+OB
2
) ...... [Since, OD=OB & OA=OC]
∴AB
2
+BC
2
+CD
2
+AD
2
=4(OA
2
+OB
2
)
solution
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