$(1)A\frac{*}{b}ar(ABC)=\frac{A}{b}ar(BC)+\frac{A}{b}ar(B)C+A\frac{B}{b}ar(C)$

$(2)\frac{A}{b}ar(B)+\frac{B}{b}ar(C)\frac{+}{b}ar(A)C\frac{?}{b}=ar(A)B\frac{+}{b}ar(B)C+\frac{A}{b}ar(C)$

$(4)AB+\frac{A}{b}ar(B)\frac{+}{b}ar(A)B\frac{+}{b}ar\frac{A}{b}ar(B)=1$

$(5)\frac{A}{b}ar(B)+BD+$ADE$+\frac{D}{b}ar(A)=\frac{A}{b}ar(B)+D$

$(6)AB+AC+\frac{A}{b}ar\frac{B}{b}ar(C)=A$

$(7)AB+\frac{A}{b}ar(C)+\frac{A}{b}ar\frac{B}{b}ar(C)=AB+\frac{A}{b}ar(C)$

$(8)AB\frac{+}{b}ar(A)C\frac{+}{b}ar(B)C=AB+C$

$(9)\frac{A}{b}ar(B)\frac{+}{b}ar(ACD)+B\frac{+}{b}ar(C)\frac{+}{b}ar(D)=1$

$(10)AB+\frac{A}{b}ar(C)\frac{+}{b}ar(B)C\frac{+}{b}ar(C)B\frac{+}{b}ar(B)D\frac{+}{b}ar(D)B+$ADEF$=A\frac{+}{b}ar(B)D+\frac{C}{b}ar(D)+\frac{B}{b}ar(C)$