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M and N are points on opposite sides AD and BC of a parallelogram ABCD such that MN passes through the point of intersection O of its diagoinals AC and BD. Show that MN is bisected at O.

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Solution :
`triangle AOM ~=triangle CON` <br> ` [because angle MAO = angle NCO, AO= CO, angle MOA = angle NOC]` <br> [Note Diagonals of a ||gm bisect each other]. <br> `therefore MO=NO " (c.p.c.t.)." ` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RSA_MATH_IX_C10_E03_004_S01.png" width="80%">

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