Do you think Betty and Daniel are going to fall in love someday? Would you like them to?
2007-02-12 23:28:52 · 14 answers · asked by Jenny 1 in Entertainment & Music ➔ Polls & Surveys
As far as i can remember, they ended up liking each other and she changed from an ugly duckling to a beautiful swan...
2007-02-13 00:53:48 · answer #1 · answered by kim B 4 · 1⤊ 0⤋
No, I don't see that happening. I think they are from two different worlds and it would never work out. However, I do think Betty will slowly evolve over the years into a "swan". So, maybe in later seasons that could happen. Still, I like Betty just the way she is now!
2016-05-24 04:52:29 · answer #2 · answered by Anonymous · 0⤊ 0⤋
They are good for each other, but what about that other dude Betty used to be with? They are a good match too. I guess it's going to be a love triangle.
2007-02-12 23:31:16 · answer #3 · answered by DARIA. - JOINED MAY 2006 7 · 1⤊ 1⤋
I would like them to get together and fall in love. They need each other.
2007-02-12 23:35:25 · answer #4 · answered by Bazinga 7 · 0⤊ 0⤋
i reckon he'll see her for the real her and they'll at least go on a date together. i hope they do she deserves it, shes such a giving,kind,nice person. but he can b a jerk sometimes. go betty
2007-02-12 23:33:11 · answer #5 · answered by rebecca a 1 · 0⤊ 0⤋
I thought so in the beginning but he was kissing her sister last week. I'd be happy if she ended up with accountant now they were cute.
2007-02-12 23:31:52 · answer #6 · answered by Anonymous · 1⤊ 0⤋
I don't care if they do fall in love. If they do then I hope they are happy.
2007-02-12 23:31:47 · answer #7 · answered by icunurse85 7 · 0⤊ 0⤋
they probly will do, they all end up being like her and not caring about appearances
2007-02-12 23:31:48 · answer #8 · answered by cinders162002 3 · 1⤊ 0⤋
I think they will and i'd like them too.
2007-02-12 23:40:26 · answer #9 · answered by Anonymous · 0⤊ 0⤋
I actually think so..think that would be soooo romantic
2007-02-12 23:31:21 · answer #10 · answered by mousecurtain 2 · 0⤊ 0⤋