Angle A+ Angle ABC+ Angle ACB= 180.

Then, the angle in the triangle ABC is above. Now look at the other one. Let the bisector CF and AB intersect at point m, and in the triangle BCM, we can get:

Angle CMB+ Angle ABC+ Angle BCM= 180.

Because the angle BCM= 1/2 and the angle ACB, it can be concluded that:

Angle CMB= 1/2 Angle ACB+ Angle A.

Now look at the next triangle, which is the triangle MDF, and you can get:

Angle CMB+ Angle MDF+ Angle MFD= 180.

Observation shows that angle MDF= angle ABC, and the above formula is equivalent to:

1/2 angle ACB+ angle A+ angle ABC+ angle MFD= 180.

Because:

Angle A+ Angle ABC+ Angle ACB= 180.

Therefore, it can be concluded that:

Angle ACB= 1/2 Angle ACB+ Angle MFD

That is, angle MFD= 1/2 angle ACB= angle ECF.

And because angle MFD= angle CFE (diagonal theorem), so:

Angle CFE= angle ECF

That is, EF=EC, so that DF=DB can also be obtained in the same way, that is, the problem becomes to find AB+AC=? If your condition BC=4cm becomes AC=4cm, then the answer comes out now.

Of course, what I said is very complicated, because I'm afraid I know something that others don't, and there's a lot of nonsense. Please forgive me. . .