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

Question

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

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Aubrey 1 week 2021-09-09T17:51:59+00:00 2 Answers 0 views 0

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    0
    2021-09-09T17:52:59+00:00

    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.

    0
    2021-09-09T17:53:02+00:00

    So here is your answer….

    AC = BD

    OA=OC

    OB=OD

    and angle AOB 90′

    Hence The diagonals of a square are equal in length and the diagonals are bisect each other…

    I hope that it gonna help you

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