In $\Delta ABC$,$D$ is any point on the side $BC$. Prove that the perimeter of $\Delta ABC > 2 AD$.
(N/A) In $\Delta ABD$,by the triangle inequality theorem,the sum of any two sides is greater than the third side: $AB + BD > AD$ (Equation $1$).
In $\Delta ACD$,by the triangle inequality theorem: $AC + CD > AD$ (Equation $2$).
Adding Equation $1$ and Equation $2$,we get: $(AB + BD) + (AC + CD) > AD + AD$.
Rearranging the terms: $AB + AC + (BD + CD) > 2 AD$.
Since $D$ is a point on $BC$,$BD + CD = BC$.
Therefore,$AB + AC + BC > 2 AD$.
This shows that the perimeter of $\Delta ABC$ is greater than $2 AD$.
In $\Delta ACD$,by the triangle inequality theorem: $AC + CD > AD$ (Equation $2$).
Adding Equation $1$ and Equation $2$,we get: $(AB + BD) + (AC + CD) > AD + AD$.
Rearranging the terms: $AB + AC + (BD + CD) > 2 AD$.
Since $D$ is a point on $BC$,$BD + CD = BC$.
Therefore,$AB + AC + BC > 2 AD$.
This shows that the perimeter of $\Delta ABC$ is greater than $2 AD$.
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