Question

ABC is a triangle <br />prove that:-<br />1.AB+BC>AC<br />2.BC+AC>AB<br />3.AB+AC>BC<br /><br />theoram:-the sum of two sides of a is greater than the third side.

Answer 1

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AM is a median of a triangle ABC. Is AB+BC+CA>2AM?

In triangle ABC, AM is a median.

In triangle ABM, AB+BM>AM… (1) [the sum of two sides of a triangle are greater than the third side].

In triangle ACM, AC+CM>AM… (2) [the sum of two sides of a triangle are greater than the third side].

Add (1) and (2) to get

AB+BM+AC+CM>AM+AM, or

AB+(BM+CM)+AC>2AM, or

AB+BC+CA>2AM. Proved

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the expression is hence true.

Note: The original question says the median is AM but I failed to make note of it and solved with AD as median. But nevertheless the solution remains.