∴ef:ad=bf:bd=cg:cd=(bf+cg):(bd+cd)=(bc-fg):bc

That is x: 16=(30-y/x):30.

y=(- 15/8)x^2+30x

The minimum value of x is 0 and the maximum value is AD (excluding 0 and AD).

The value range of ∴x is (0, 16).

2.( 1) proof: ∴OP:BD=AO:AD.

OQ‖AC,∴OQ:AC=DO:AD

∴op:bd+oq:ac=(ao+do):ad= 1

(2)AC⊥BD, then OP⊥OQ, y=OP*OQ/2.

OQ=x，OP =( 1-OQ/AC)* BD = 12-x

∴y=x( 12-x)/2

The minimum value of x is 0 and the maximum value is AC (excluding 0 and AC).

The value range of ∴x is (0, 12).

3.BC= 10, area =30, height = 30× 2/ 10 = 6.

From 1 topic, DP:AH=(BC-DG):BC (see 1 topic for the proof method).

DP =(BC-DG)* AH/BC =( 10-x)×6/ 10 = 6-(3/5)x

y=dp*dg=[6-(3/5)x]×x=6x-(3/5)x^2

The minimum value of x is 0 and the maximum value is BC (excluding 0 and BC).

The value range of ∴x is (0, 10).