Question:
M is a point on side BC of a triangle ABC such that AM is the bisector of ∠BAC. Is it true to say that perimeter of the triangle is greater than 2AM? Give
reason for your answer?
Solution:
Yes, in $\triangle A B C, M$ is a point of side $B C$ such that $A M$ is the bisector of $\angle B A C$.
$\ln \triangle A B M, \quad A B+B M>A M \quad \ldots$ (i)
[sum of two sides of a triangle is greater than the third side]
$\ln \triangle A C M$ $A C+C M>A M$ ... (ii)
[sum of two sides of a triangle is greater than the third side]
On adding Eqs. (i) and (ii), we get
$(A B+B M+A C+C M)>2 A M$
$\Rightarrow \quad(A B+B M+M C+A C)>2 A M$
$\Rightarrow \quad A B+B C+A C>2 A M \quad[\because B C=B M+M C]$
$\therefore \quad$ Perimeter of $\triangle A B C>2 A M$