Expand AD to e, make ED=AD, connect BE,

In △BDE and △CDA,

BD=CD，∠BDE=∠CDA，ED=AD，

∴△BDE≌△CDA(SAS)，

∴BE=AC，

In △ Abe,

∵ ab+be > AE (the sum of two sides of a triangle is greater than the third side),

∴AB+AC>2AD

2、

BE=AC=3，AB=5

In △ Abe,

∫a B- BE < AE < a b+BE，

∴5-3