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| Theorem 5 |
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| The diagonals of a rhombus are perpendicular to each other. |
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| ABCD is a rhombus. Diagonal AC and BD intersect at O. |
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| AC and BD bisect each other at right angles. |
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| A rhombus is a parallelogram such that |
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| AB=DC=AD=BC ---(i) |
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| Also the diagonals of a parallelogram bisect each other. |
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| Hence BO=DO and AO=OC ---(ii) |
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| Now compare triangles AOB and AOD, |
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| AB=AD (from (i) above) |
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| BO=DO (from (ii) above) |
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| AO=AO (common side) |
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 (SSS congruency condition) |
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 (corresponding parts of corresponding parts) |
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| BD is a straight line segment. |
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| i.e., the diagonals bisect at right angles. |
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| Hence the theorem is proved. |
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