Answers

2015-11-03T12:22:48+05:30
GIVEN - A SQUARE ABCD AND DIAGONALS AC AND BD BISECT AT O.
TO PROVE - AC = BD,
                    OA=OC, OB =OD
                   ∠AOB =90°
PROOF -
1) IN ΔABC AND ΔDCB
          AB=DC ( EQUAL SIDES OF SQUARE)
         ∠ABC=∠DCB (90° EACH)
         BC = CB(COMMON SIDE)
    ∴ΔABC ≡ ΔDCB (SAS RULE)
          AC=DB ( CPCT)
2) IN ΔAOB AND ΔCOD
         ∠AOB=∠COD (VERTICALLY OPPOSITE ANGLES )
         ∠ABO=∠CDO (ALTERNATE INTERIOR ANGLES )
          AB = CD ( EQUAL SIDES OF SQUARE )
    ΔAOB ≡ ΔCOD  ( AAS RULE )
      AO=CO AND OB = OD ( CPCT)
3) IN ΔAOB AND ΔCOB
         AO = CO(PROVED ABOVE)
         AB=CB ( EQUAL SIDES OF SQUARE)
         BO=BO (COMMON)
   ΔAOB≡ΔCOB ( SSS RULE)
     ∴∠AOB=∠COB ( CPCT )
    ∠AOB+∠COB=180° ( LINEAR PAIR )
      2∠AOB=180° (SINCE,∠AOB=∠COB)
       ∠AOB=180°/2
       ∠AOB=90°

THEREFORE, PROVED THAT IN A QUAD. IF DIAGONALS ARE EQUAL AND THEY PERPENDICULARLY BISECT EACH OTHER THEN THEY THE QUAD. IS A SQUARE.


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