# A square ABCD has side AB=1, and K is between B and C. E is on the extension of AK, with K between A and E. Lines EC and AD intersect at F. If the area of triangle AEF=3 and AE: AK=7, compute BK.

The length of segment BK is 13/49. We are given a square ABCD with AB=1. Point K is located between B and C, with point E on the extension of AK such that K is between A and E. Lines EC and AD intersect at F.

The area of `Delta AEF` is 3, and `"AE":AK"=7` . We are asked to find the length of the segment BK.

First note that if K is between B and C, this means that K lies on `bar(BC)` and BK+KC=BC. Also, K lies on `bar(AE)` with AK+KE=AE.

**See attachment for diagram **

We know that `Delta AEF " is similar to " Delta KEC " by " A A " similarity"` The scale factor is 7:6 from the given AE:AK=7. (Let AK=x; then AE=7x and KE=6x, so AE:KE=7:6.)

From E, drop perpendicular EG to the extension of AD. Let H be the intersection of EG and BC.` `

Triangle GEA is similar to triangle BAK so GE/BA=AE/AK=7/1 (given); with BA=1, we have GE=7.

The area of triangle AEF is 3, so 1/2(AF)(EG)=3 ==> AF(7)=6 so AF=6/7. Then, KC/AF = 6/7 ==> KC=36/49.

BK=BC-KC=1-36/49=13/49.

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