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Question
Prove by vector method that the diagonals of a rhombus bisect each other at right angles
Solution
Let ABCD be a rhombus
To prove `bar"AC" * bar"BD"` = 0
We have AB = BC = CD = DA
Now `bar"AC" = bar"AB" + "BC"`
`bar"BD" = bar"BC" + bar"CD"`
= `bar"BC" - bar"AB"` .......`("Since" bar"CD" = - bar"AB")`
`bar"AC" * bar"BD" = (bar"BC" + bar"AB") * (bar"BC" - bar"AB")`
= `(bar"BC")^2 - (bar"AB")^2`
= `("BC")^2 - ("AB")^2`
`bar"AC" * bar"BD"` = 0
∴ `bar"AC"` ⊥' to `bar"BD"`
Hence the result.
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