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

Explanation

Proof:
(i) In a `/_\ABC and /_\BAD`,
`AB=AB` ( common line)
`BC=AD` ( opppsite sides of a square)`/_ABC=/_BAD ( =90^0` )
`/_\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)
Similarly OB=OD ----------(ii)
From (i) and (ii) AC and BD bisect each other.
Now in a ΔOBA and ΔODA,
OB=OD ( from (ii) )
BA=DA
OA=OA ( common line)
`/_\AOB=/_\AOD`----(iii) ( by CPCT
`/_AOB+/_AOD=180^0` (linear pair)
`2/_AOB=180^0`
`/_AOB=/_AOD=90^0`
∴AC and BD bisect each other at right angles.