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In the given triangle ABC, D is any point on side BC, then: AB+BC+AC>2AD
A.True
B.False

Answer
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Hint: We are going to solve the given problem using the triangular inequality of triangles. It gives the information that the sum of any two sides of a triangle is greater than the third side.

It is given in a triangle ABC, D is a point anywhere on BC. So when we put the point D on BC and joins that point from vertex A, triangle ABC will divide into two triangles, triangle ABD, triangle ACD
In triangle ABD,
AB+BD>AD [because, the sum of any two sides of a triangle is always greater than the third side]
In triangle ADC,
AC+DC>AD [because, the sum of any two sides of a triangle is always greater than the third side]
Adding these we get,
AB+BD+AC+DC>AD+AD
$ \Rightarrow $ AB + (BD + DC) + AC > 2AD
$ \Rightarrow $ AB + BC + AC > 2AD
Hence AB+BC+AC>2AD

Note: The triangle inequality theorem states that the sum of any two sides of a triangle must be greater than the measure of the third side. It is applicable on all three sides of any triangle.