∵△ABC∽△A'B'C'
又AD⊥BC,A'D'⊥B'C'
∴设AB:A'B'=AD:A'D'=BC:B'C'=k
∵S△ABC=1/2*BC*AD,S△A'B'C'=1/2*B'C'*A'D'
∴S△ABC:S△A'B'C'=1/2*BC*AD:1/2*B'C'*A'D'=k^2
即S△ABC:S△A'B'C'=(AB:A'B')^2
∵△ABC∽△A'B'C'
又AD⊥BC,A'D'⊥B'C'
∴设AB:A'B'=AD:A'D'=BC:B'C'=k
∵S△ABC=1/2*BC*AD,S△A'B'C'=1/2*B'C'*A'D'
∴S△ABC:S△A'B'C'=1/2*BC*AD:1/2*B'C'*A'D'=k^2
即S△ABC:S△A'B'C'=(AB:A'B')^2