$M$ is a point on side $BC$ of a triangle $ABC$ such that $AM$ is the bisector of $\angle BAC$. Is it true to say that the perimeter of the triangle is greater than $2\, AM$? Give reason for your answer.
(A) We have to prove that $AB + BC + AC > 2\, AM$.
According to the triangle inequality theorem,the sum of any two sides of a triangle is greater than the third side.
In $\triangle ABM$,we have:
$AB + BM > AM$ $\ldots (1)$
In $\triangle ACM$,we have:
$AC + CM > AM$ $\ldots (2)$
Adding equations $(1)$ and $(2)$,we get:
$AB + BM + AC + CM > AM + AM$
$AB + AC + (BM + CM) > 2\, AM$
Since $M$ is a point on $BC$,$BM + CM = BC$. Substituting this into the inequality:
$AB + AC + BC > 2\, AM$
Thus,the perimeter of the triangle $ABC$ is indeed greater than $2\, AM$.
According to the triangle inequality theorem,the sum of any two sides of a triangle is greater than the third side.
In $\triangle ABM$,we have:
$AB + BM > AM$ $\ldots (1)$
In $\triangle ACM$,we have:
$AC + CM > AM$ $\ldots (2)$
Adding equations $(1)$ and $(2)$,we get:
$AB + BM + AC + CM > AM + AM$
$AB + AC + (BM + CM) > 2\, AM$
Since $M$ is a point on $BC$,$BM + CM = BC$. Substituting this into the inequality:
$AB + AC + BC > 2\, AM$
Thus,the perimeter of the triangle $ABC$ is indeed greater than $2\, AM$.
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