show that the diagonals of a square are equal and bisects eachother at right angle triangle
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In square ABCD AC and DB are daingonal intersecting at o prove that triangle AOB =~triangle BOC .: by. Cpct AO =CO :. Diagonal are bisected and angle AOB =angle BOC but they are linear pair .: angle AOB +angle BOC =180
2angle BOC = 180
Angle BOC =180/2
Angle BOC =90
:.proved
2angle BOC = 180
Angle BOC =180/2
Angle BOC =90
:.proved
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Step-by-step explanation:
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.
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