Question

In the adjacent figure, D is any point on the side BC of $△ABC$. Show that $AB+BC+CA>2AD$.

Answer 1

Triangle inequality: The sum of two sides will always be greater than the third side.

In $△ABD$,

According to the triangle's inequality, the sum of two sides will always be greater than the third side.

$AB+BD>AD------------\left(i\right)$

In $△ADC$,

According to the triangle's inequality, the sum of two sides will always be greater than the third side.

$AC+CD>AD------------\left(ii\right)$

Adding equations (i) and (ii),

$$\begin{gathered} AB+BD+AC+CD>AD+AD \\ AB+AC+BD+CD>2AD \\ AB+AC+BC>2AD-------(BD+CD=BC) \end{gathered}$$

Hence proved.