Trapezoid: If AB = (2/3)DC, prove that AE = (3/5)AB + (2/5)AD.

[Plavergne]

I have been stuck on this question all day. Help Please!

ABCD is a trapezoid whose diagonals AC and BD intersect at the point E. If AB = (2/3)DC, prove that AE = (3/5)AB + (2/5)AD.

 

[ksdhart2]

I have been stuck on this question all day.

Great! That means you ought to have loads of work to share with us, as per the rules laid out in the Read Before Posting thread that's stickied at the top of each sub-forum (you did read it, right? ). Please share with us all of the work you've done on this problem, even the parts you know for sure are wrong. Thank you.

 

[Plavergne]

[FONT=&quot]Let AB = y
Let DC = x

AC = AD+DC = y+1.5x
[/FONT]AE = h(AC)
AE = h(y+1.5x)
[FONT=&quot]BD = BA+AD
= −AB + AD
=-x+y
[/FONT]BE = g(BD)
BE = g(-x +y)

[FONT=&quot]But AB+BE+EA = 0
Therefore:
x +[/FONT]h(y+1.5x) + g(-x +y) = 0
x + hy +1.5hx +gy -gx = 0
x (1 +1.5h -g) + y (h+g) = 0

Then I'm stuck
[FONT=&quot]

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