Can Black reach full equality in this game? I have played this game as Black against computer( difficulty was set to HARD ). I was trying to test Alatortsev line I found in ECO D, 1987 that claimed Black was equal:

[Title "Taken from ECO D ( published 1987 )"]
[fen ""]

1.d4 d5 2.c4 e6 3.Nc3 Be7 4.cxd5 exd5 5.Bf4 c6 6.Qc2 Bf6 7.e3 Ne7 8.Bd3 g6= 


saying "with the idea of Bf5".

After g4 my position became barely tenable. 

Is there something I could have done better in order to reach full equality? 

How does modern theory evaluate 8...g6?

Here is the game:

[White "Shredder"]
[Black "AlwaysLearningNewStuff"]
[fen ""]

1.d4 d5 2.c4 e6 3.Nc3 Be7 4.cxd5 exd5 5.Bf4 c6 6.Qc2 Bf6 7.e3 Ne7 8.Bd3 g6= 9.Nf3 Bf5 10.Bxf5 Nxf5 11.g4! $40 Ne7 12.Qb3 Qb6 13.Qa3! Na6 14.g5 Bg7 15.O-O-O Qb4! $7 16.Qxb4 Nxb4 17.a3 Na6 18.e4! $16 $40 O-O-O 19.Be5! Bxe5 20.Nxe5 Rhf8 21.exd5 Nxd5 22.Ne4! Nac7 23.Rhe1 Ne6 24.h4 Kc7 25.Nc5?! Nxc5 26.dxc5 f6!= 27.Nd3 Rde8 28.Kc2 Kd8 29.Rxe8+ Rxe8 30.Nf4 fxg5 31.Nxd5 cxd5 32.Rxd5+ Kc7 33.hxg5 Re2+ 34.Kc3 Rxf2 35.b4 Rf3+ 36.Rd3 Rf5 37.Rh3 Rxg5!= 38.Rxh7+ Kc6 39.Rh8 a5 40.Kb3 axb4 41.axb4 b6 42.cxb6 Kxb6 43.Rb8+ Kc6  44.Kc3 Rb5 45.Rc8+ Kb6 46.Kc4 Rf5 47.Rb8+ Kc6 48.b5+ Kc7 49.Ra8 g5 50.Ra7+ Kb6 51.Ra6+ Kb7 52.Rg6 Re5 53.Kb4 Rf5 54.Ka5 Re5 55.Rg7+ Kb8 56.Ka6 Re6+ 57.b6 Re8 58.Rb7+ Ka8 59.Rg7 Kb8 60.Rb7+ Ka8 61.Rg7 Kb8 62.Rb7+ 1/2-1/2
Deep Rybka's 4 opening book (with pre-calculated scores on positions in it) gives the position resulting by

    1.d4 d5 2.c4 e6 3.Nc3 Be7 4.cxd5 exd5 5.Bf4 c6 6.Qc2 Bf6 7.e3 Ne7 8.Bd3 g6


a +0.52 score, so definitely some advantage for White. It contains the following continuations:

[fen ""]

1. d4 d5 2. c4 e6 3. Nc3 Be7 4. cxd5 exd5 5. Bf4 c6 6. Qc2 Bf6 7. e3 Ne7 8. Bd3 g6 9. Nf3 Bf5 10. h4 (10. O-O Bxd3 11. Qxd3 O-O 12. h4 (12. b4 Nd7))


and several sub variations.