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

Expert Answer:

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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