Mathematics
In △ABC, D is any point on BC. Prove that : AB + BC + AC > 2 AD.

Triangles
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Answer
We know that,
In a triangle, sum of any two sides is always greater than the third side.
In △ABD,
⇒ AB + BD > AD …..(1)
In △ACD,
⇒ AC + CD > AD …..(2)
Adding eq.(1) and (2), we have :
⇒ AB + BD + AC + CD > AD + AD
⇒ AB + (BD + CD) + AC > 2AD
⇒ AB + BC + AC > 2AD.
Hence, proved that AB + BC + AC > 2AD.
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