Prove that in a square diagnols are equal and bisect each other at right angles
Asked by 100.akash
| 21st Sep, 2008,
03:42: PM
let ABCD be a square of side 'a'. and AD and BC are its diagonals.
then we know that AC2 = AB2+BC2 = 2a2
similarly BD2 = BC2+CD2 = 2a2
so the length of diagonals are same and equal to 2a
now in vector we can write AC = AB + BC and BD = BA + AD
now AC.BD = (AB + BC) . (BA + AD) = AB.BA + AB.AD + BC.BA + BC.AD
AB.BA =-AB.AB = -|AB|2 = -a2 and AB.AD = 0(since AB AD )
similarly BC.BA = 0 and BC.AD = a2 (since BC AD )
so AC.BD = -a2 +0+0+a2 =0
which means AC and BD are perpendicular to each other.
Answered by
| 21st Sep, 2008,
04:53: PM
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