# in the diagram below of triangle ADE, B is a point on line AE, and C is a point on line AD such that line bc is parallel to line ed, ac=x-3, be=20, ab=16, and ad= 2x+2. find the length of line ac....

in the diagram below of triangle ADE, B is a point on line AE, and C is a point on line AD such that line bc is parallel to line ed, ac=x-3, be=20, ab=16, and ad= 2x+2. find the length of line ac.

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Since `bar(bc)||bar(ed)` we have `Delta ACB` ~ `Delta ADE` by AA~.

Then `(AC)/(AD)=(AB)/(AE)`

Substituting the known values we get:

`(x-3)/(2x+2)=16/36` Simplifying `16/36=4/9` we solve for x using the means-extremes product property of proportions:

`4(2x+2)=9(x-3)`

`8x+8=9x-27`

`x=35`

**Then `AC=x-3=32` **

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We can check this in a number of ways. There is a theorem that states that a line parallel to one side of a triangle that intersects the other two sides of the triangle divide the other sides proportionally.

And indeed we see that `16/32=20/40`