Line AB extends to C, that is, BC=2AB, D is the midpoint of line AB, E and F are points on BC, BE:EF= 1:2, EF:FG=2: 5, AC=60CM. The long answer to DE is BC=2AB, DB = 1.

DB = 1/4BC = 1/6AC = 10cm。

So BC=40cm.

And because be: ef =1:2, ef: fc = 2: 5, BC=BE+EF+FC, then

BC=BE+2BE+4/5BE= 19/5BE

So BE=200/ 19cm.

EF=2BE=400/ 19cm。

Therefore, de = db+be =10+200/19 = 390/19 cm.

DF = DE+EF = 390/ 19+400/ 19 = 790/ 19cm

Given the line segment AB= 10cm, C is the midpoint of AB, D is a point above AC, and DC:DB=2:7, find the length of line segment AD.

Answer: There are two possibilities.

___________________

Post-ad century

Let CD=2a, then DB=7a.

BC=DB-CD=7a-2a=5a

You Athena Chu AB= 10,

So BC=5

So 5a=5,

a= 1

So AD= 10-DB= 10-7a=3.

or

___________________

A C D B

10/2=5

2+7=9

CD= 5* 2/9= 10/9

AD= 10/9+5=55/9