Question

In a$△ABC$, $AD$is median. Show that $AB+BC+AC>2AD$

Answer 1

Given, $AD$is median in $△ABC$

To prove, $AB+BC+AC>2AD$

Let $△ABC$where $AD$is a median

In$∆ABD$, by the property of the triangle we have

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

In$∆ACD$, by the property of the triangle we have

$AC+DC>AD......\left(ii\right)$

Adding $\left(i\right)and\left(ii\right)$ we get

$$\begin{gathered} AB+BD+AC+DC>AD+AD \\ AB+BC+AC>2AD(\sin ceBD+DC=BC) \end{gathered}$$

Hence, $AB+BC+AC>2AD$ proved.