In the diagram, square ABCD has sides of length 4, and triangle ABE is equilateral. Line segments BE and AC intersect at P. Point Q is on BC so that PQ is perpendicular to BC and PQ=x. Compute AP.
PQ = CQ = x ∠PCQ = 45º ∠PBQ = 30º
tan(45º) * x = tan(30º) * (4 - x)
x = 2(√3 -1) ≈ 1.464