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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