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The diagonals of a parallelogram bisect ...

The diagonals of a parallelogram bisect each other.

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`ABCD` is a parallelogram, diagonals `AC` and `BD` intersect at `O`
In `triangleAOD` and `triangleCOB`,
`angleDAO=angleBCO` (alternate interior angles)
`AD=CB`
`angleADO=angleCBO` (alternate interior angles)
`triangleAOD~=triangleCOB` (By `ASA`)
Hence, `AO=CO` and `OD=OB` (By `CPCT`)
Thus, the diagonals of a parallelogram bisect each other.
Hence proved.
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Knowledge Check

  • If A (1,-6) , B= (5,-2) and C = (12 , -9) are the three consecutive vertices of a parallelogram ,then the find fourth vertex . The following are the steps involved in solving the above problem . Arrange them in sequential order from beginning to end. (A) (5 +x)/(2) = (13)/(2) , (-2 +y)/(2) = (-15)/(2) implies x = 8 and y = -13 . Therefore , D = (8 , -13) . (B) therefore ((5 +x)/(2) , (-2+y)/(2)) = ((1+12)/(2) , (-6-9)/(2)) (C) Let the fourth vertex be D = (x , y) . (D) We know that diagonals of a parallelogram bisect each other.

    A
    ACBD
    B
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    C
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    D
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  • If diagonals of a parallelogram ABCD intersect each other in M, then bar(OA) + bar(OB) + bar(OC) + bar(OD) =

    A
    `bar(OM)`
    B
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    3`bar(OM)`
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