$\mathrm{\angle}BCD=\mathrm{\angle}BCA+\mathrm{\angle}2$ [ Since, $\mathrm{\angle}1=\mathrm{\angle}2$ ]

$\therefore $ \angle BCD>\angle 2

Now, in $\mathrm{\u25b3}BCD,$

$\Rightarrow $ \angle BCD>\angle 2

$\Rightarrow $ BD>BC [ Side opposite to the larger angle is longer ]

$\Rightarrow $ AB+AD>BC [ Since, $BD=AB+AD$ ]

$\Rightarrow $ AB+AC>BC [ Since, $AD=AC$ by construction ]

Similarly, we can prove

AB+BC>AC

And AC+BC>AB

$\therefore $ The given statement is true.