∴

DE=3- 1=2

2. There are three situations:

⑴

Intersect with AB

Prove that △ ACD △ BCE

DE=∣a-b∣

⑵

1. If it coincides with AC or BC, then DE=AC=BC.

(3) 4. It does not intersect with AB, as shown in the figure.

Create CF⊥ι

Then: AD‖CF‖BE

∠ACD=BFC(90？ -∠ACF)=∠CBE

Similarly: ∠DAC=∠BCE

∴△ACD≌△BCE

DE=a+b