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.
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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
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A C D B
10/2=5
2+7=9
CD= 5* 2/9= 10/9
AD= 10/9+5=55/9