defiantly lancome! they do amazing mascaras! i always use it, Ive had diorshow and it clumped my lashes really bad and its hard to use as the brush is HUGE
2007-01-22 04:08:41
·
answer #1
·
answered by Anonymous
·
0⤊
0⤋
I can't tell you which is better because I don't know anything about DiorShow. But I can tell that I've been using Lancome Definicils (non-waterproof version) for a long time and I love it. I prefer it over other mascaras. My eyelashes look smooth and soft, and not hard and messy. ;)
If you have any question, let me know. ;)
2007-01-22 03:42:18
·
answer #2
·
answered by yahaks 1
·
1⤊
0⤋
the two mascaras are stunning of their own top... yet once you choose for quantity, decide for DiorShow. that's a cult famous and that i take advantage of it on maximum of my clientele. although, its lengthening and defining understand-how are short-lived. For length and definition, Lancome Definicils is the thank you to pass. probable between the wonderful mascaras I even have ever used. attempt them the two; you are the marvelous choose :)
2016-10-31 23:52:55
·
answer #3
·
answered by ? 4
·
0⤊
0⤋
so am i totaly am in love with makeup. i have chanel mascara and never tried Dior or Lancome. but i would have to do with Dior cause it's alot nicer. Good Luck =]
<33 emily
2007-01-22 05:22:53
·
answer #4
·
answered by *Dancer* 2
·
0⤊
0⤋
I've heard Diorshow is a great mascara, albeit pricey.
2007-01-22 03:06:19
·
answer #5
·
answered by Last Call 4
·
0⤊
0⤋
Defincils or Hypnose by Lancome
2007-01-22 02:24:20
·
answer #6
·
answered by Kate M 4
·
0⤊
0⤋
Use Dior Show Mascara i is very nice.
2007-01-22 22:28:49
·
answer #7
·
answered by geetajanodimahadevigurujanodikan 1
·
0⤊
0⤋
Dior
2007-01-22 04:18:50
·
answer #8
·
answered by Anonymous
·
0⤊
0⤋
DIOR, and you can get the same thing from Benifit for a few dollars less
2007-01-22 02:24:55
·
answer #9
·
answered by Juleette 6
·
0⤊
0⤋
Dior is better to me, but maybe you could try it yourself first (if testers are available)
2007-01-22 02:24:57
·
answer #10
·
answered by Vesna G 5
·
0⤊
0⤋