Show that the diagonals of a square are equal and bisect each other at right angles.

Explanation

(i)In a `/_\ABC \and\ /_\BAD`
`AB=AB`(common line)
`BC=AD`(opppsite sides of a square)
`/_ABC=/_BAD = 90^@`
`/_\ABC~=/_\BAD`( By SAS property)
`AC=BD` ( by CPCT).
(ii) In a,`/_\OAD and /_\OCB`
`AD=CB`(opposite sides of a square)
`/_OAD=/_OCB`(transversal AC)
`/_ODA=/_OBC`(transversal BD)
`/_\OAD~=/_\OCB` (ASA property)
`OA=OC` (i)
`OB=OD`(ii)
From (i) and (ii) AC and BD bisect each other.
Now in a `/_\OBA \and\ /_\ODA,OB=OD`(from (ii))
`BA=DA`
`OA=OA`(common line)
`/_\AOB=/_\AOD`(iii)(by CPCT)
`/_AOB+/_AOD=180^@` (linear pair)
`2/_AOB=180^@`
`/_AOB=/_AOD=90^@`
AC and BD bisect each other at right angles.