Show that the diagonal of a …

Given that ABCD is a square. To prove : AC=BD and AC and BD bisect each other at right angles. Proof:  (i)  In a ΔABC and ΔBAD, AB=AB ( common line) BC=AD ( opppsite sides of a square) ∠ABC=∠BAD ( = 90° ) ΔABC≅ΔBAD( By SAS property) AC=BD ( by CPCT). (ii) In a ΔOAD and ΔOCB, AD=CB ( opposite sides of a square) ∠OAD=∠OCB ( transversal AC ) ∠ODA=∠OBC ( transversal BD ) ΔOAD≅ΔOCB (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 ) ΔAOB=ΔAOD----(iii) ( by CPCT ∠AOB+∠AOD=180°  (linear pair) 2∠AOB=180° ∠AOB=∠AOD=90° ∴AC and BD bisect each other at right angles.