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In Figure, l \parallel m and M is the mid-point of a line segment AB. Show that M is also the mid-point of any line segment CD, having its endpoints on l and m, respectively.

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Given : l \parallel m

M is the mid-point of AB then AM = MB

To show:- M is the mid-point of CD.

Proof:- l \parallel m

Then, In \triangleACM and \triangleDMB

\angleACM = \angleBDM        (Alternate interior angles are equal)

\angleAMC = \angleDMB        (vertically opposite angles are equal)

AM = MB                   (given)

then by AAS criterion of congruence

\triangleACM \cong \triangleBDM

Then DM = CM          (CPCT)

and CD = DM + MC = 2 DM

DM = CM = CD

Then M is also the mid-point of CD.

Hence proved.

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