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# If $ \mathrm{B} $ is the mid point of $ \overline{\mathrm{AC}} $ and $ \mathrm{C} $ is the mid point of $ \overline{\mathrm{BD}} $, where $ \mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D} $ lie on a straight line, say why $ \mathrm{AB}=\mathrm{CD} $?

Given:

\( \mathrm{B} \) is the mid point of \( \overline{\mathrm{AC}} \) and \( \mathrm{C} \) is the mid point of \( \overline{\mathrm{BD}} \), where \( \mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D} \) lie on a straight line.

To do:

We have to say why \( \mathrm{AB}=\mathrm{CD} \).

Solution:

From the figure,

B is the midpoint of $AC$.

This implies,

$AB = BC$..........(i)

C is the midpoint of BD.

This implies,

$BC = CD$............(ii)

From (i) and (ii), we get,

$AB = CD$

Hence proved.

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