Question

4. AM is a median of A ABC. Prove that (AB + BC +CA) > 2AM. Hint. (AB+BM)> AM (In A ABM) (AC +MC) > AM (In A ACM) Now, add the two inequalities.

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Answer 1

Answer:

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Step-by-step explanation:

the answer is that it is parallel to AB

and perpendicular to DC

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Answer 2

Your correct answer is⤵️

Step-by-step explanation:

So, in the triangle ABC. We have sub triangles ABM and AMC. Using the inequality of the triangle that the sum of any two sides is always greater than or equal to the third side. Hence AB+BC+CA>2AM is proved to be true.