Question
$\text{ABCD}$ is a trapezium. Prove that:

$\text{CD} +\text{DA} +\text{AB} > \text{BC}.$
 

Answer

In $\triangle ACD,$ we have
$\text{CD} +\text{DA} > \text{CA}$
$\Rightarrow \text{CD} + \text{DA} + \text{AB} > \text{CA} +\text{AB}$
$\Rightarrow \text{CD} + \text{DA} + \text{AB} > \text{BC}. ...[\therefore \text{AB} + \text{AC} > \text{BC}]$

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