$AD$ is a median of the triangle $ABC$. Is it true that $AB+BC+CA>2AD$? Give reason for your answer.

In the given figure, $D$ and $E$ are points on side $BC$ of a $\Delta ABC$ such that $\mathrm{BD}=\mathrm{CE}$ and $\mathrm{AD}=\mathrm{AE}.$ Show that $∆\mathrm{ABD}\cong \mathrm{ACE}$.

$CDE$ is an equilateral triangle formed on a side $CD$ of a square $\mathrm{ABCD}$ (See fig.). Show that $\Delta \mathrm{ADE}\cong \Delta \mathrm{BCE}$.

In the given figure, $\mathrm{BA}\perp \mathrm{AC}, \mathrm{DE}\perp \mathrm{DF}$ such that $\mathrm{BA}=\mathrm{DE}$ and $\mathrm{BF}=\mathrm{EC}$. Show that $\Delta \mathrm{ABC}\cong \Delta \mathrm{DEF}$.