Question
Given: $\text{ED} = \text{EC}$, Prove: $\text{AB}+\text{AD}>\text{BC}$.
✓
Answer
The sum of any two sides of the triangle is always greater than the third side of the triangle.
ln $\triangle CEB,$
$\text{CF} +\text{EB} >\text{ BC}$
$\Rightarrow \text{DE} + \text{EB} > \text{BC} ...[\text{ CE} = \text{DE} ]$
$\Rightarrow \text{DB} > \text{BC} ...(i)$
ln $\triangle ADB,$
$\text{AD}+ \text{AB} >\text{ BD}$
$\Rightarrow \text{AD} +\text{ AB} > \text{BD} >\text{ BC} ...$[ from$(i) ]$
$\Rightarrow \text{AD}+ \text{AB} > \text{BC}$
ln $\triangle CEB,$
$\text{CF} +\text{EB} >\text{ BC}$
$\Rightarrow \text{DE} + \text{EB} > \text{BC} ...[\text{ CE} = \text{DE} ]$
$\Rightarrow \text{DB} > \text{BC} ...(i)$
ln $\triangle ADB,$
$\text{AD}+ \text{AB} >\text{ BD}$
$\Rightarrow \text{AD} +\text{ AB} > \text{BD} >\text{ BC} ...$[ from$(i) ]$
$\Rightarrow \text{AD}+ \text{AB} > \text{BC}$
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