BE +12CD +12CD +I10BE + 6 AF

AD!IsS BC BC AD

+12 BF) + &c.

+15 CE] + &c.

+15 CE + &c.

+12 BF) + &e.

+ 7 AG + &e.

AE s5 —&c.

BD. —&e.

BD! — &e.

Donc AÂ2=1..…A=1. 4AB=0...B 6 AC+4 B?=— A2... 6 AC=—1...C=——.

8 AD + 12 BC = — 2 AB =0 … D=0 10 AE + 16 BD + 9C2= — 9 AC— B2....10 AE = 9 1 3 1 1

Cr DAC Sub ee

12 AL + 20 BE +21CD=—2AD—2BC=0 ds. F= 0