# 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.

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.

Last updated date: 26th Sep 2023

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