Question
$\text{ABCD}$ is a trapezium. Prove that$:$
$\text{CD} +\text{DA} +\text{AB} +\text{BC} > 2\text{AC}.$
$\text{CD} +\text{DA} +\text{AB} +\text{BC} > 2\text{AC}.$
✓
Answer
In $\triangle ABC,$ we have
$\text{AB} + \text{BC} > \text{AC} ...(i)$
In $\triangle ACD,$ we have
$\text{AD} +\text{CD} > \text{AC} ...(ii)$
Adding $(i)$ and $(ii),$ we get
$\text{AB} + \text{BC} + \text{AD} + \text{CD} > 2\text{AC}.$
$\text{AB} + \text{BC} > \text{AC} ...(i)$
In $\triangle ACD,$ we have
$\text{AD} +\text{CD} > \text{AC} ...(ii)$
Adding $(i)$ and $(ii),$ we get
$\text{AB} + \text{BC} + \text{AD} + \text{CD} > 2\text{AC}.$
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