Solution: the relationship between the three is AD=BD+CD, and the reasons are as follows:

Extend DB to point e, make BE=CD, and connect AE.

Triangle ABC is an equilateral triangle.

∴∠BAC=60，AC=AB

∫∠BDC = 120

∠BAC+∠ACD+∠BDC+∠ABD=360

∴∠ACD+∠ABD= 180

∫∠ABE+∠ABD = 180

∴∠ABE=∠ACD

In △ABE and △ACD, AC=AB, ∠ABE=∠ACD, BE=CD.

∴△ABE≌△ACD

∴AE=AD,∠BAE=∠CAD

∠∠CAD+∠BAD =∠BAC = 60

∴∠BAE+∠BAD=∠EAD=60

Delta EAD is an equilateral triangle.

∴AD=DE=BE+BD=BD+CD

That is, AD=BD+CD

Question 2: When ∠ ADC = 60, ∠ ADC = ∠ ABC = 60.

So points A, B, D and C are * * * cycles, so we can get ∠ BDC+∠ BAC = 180.

∴∠BDC= 120

Same as the first question, AD=BD+CD.