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Question
Given: ED = EC
Prove: AB + AD > BC.
Solution
The sum of any two sides of the triangle is always greater than the third side of the triangle.
ln ΔCEB,
CE + EB > BC
⇒ DE + EB > BC ...[CE = DE]
⇒ DB > BC ...(i)
ln ΔADB,
AD + AB > BD
⇒ AD + AB > BD > BC ...[From (i)]
⇒ AD + AB > BC
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