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Question
In ΔABC, D is a point in the interior of the triangle. Prove that DB + DC < AB + AC.
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Solution
In the ΔABC,
AB + AC > BC .....(∵ Sum of the two sides of triangle is always greater than third side.) ....(i)
Also, in the ΔBDC,
BD + DC > BC ....(ii)
Dividing (i) by (ii),
`"AB + AC"/"BD + DC" > "BC"/"BC"`
`"AB + AC"/"BD + DC" > 1`
AB + AC > BD + DC
i.e. BD + DC < AB + AC.
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