$AD$ is a median of the triangle $ABC$. Is it true that $AB + BC + CA > 2AD$? Give reason for your answer.
(N/A) In $\triangle ABD$,we have:
$AB + BD > AD$ $\ldots(1)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
In $\triangle ADC$,we have:
$AC + CD > AD$ $\ldots(2)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
Adding $(1)$ and $(2)$,we get:
$AB + BD + CD + AC > 2AD$
Since $AD$ is the median of $\triangle ABC$,$BD = CD$. Therefore,$BD + CD = BC$.
Substituting this into the inequality,we get:
$AB + BC + CA > 2AD$
Yes,the statement is true.
$AB + BD > AD$ $\ldots(1)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
In $\triangle ADC$,we have:
$AC + CD > AD$ $\ldots(2)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
Adding $(1)$ and $(2)$,we get:
$AB + BD + CD + AC > 2AD$
Since $AD$ is the median of $\triangle ABC$,$BD = CD$. Therefore,$BD + CD = BC$.
Substituting this into the inequality,we get:
$AB + BC + CA > 2AD$
Yes,the statement is true.
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