In the diagram, square ABCD has sides of length 4, and triangle ABE is equilateral. Line segments BC and AC intersect at P. Point Q is on BC so that PQ is perpendicular to BC and PQ=x.
Find the value of x in simplest radical form.
Notice that m∠PBQ = 90° - 60° = 30° so △PBQ is a 30-60-90 triangle. So
BQ = x√3
And notice m∠PCQ = 45° so △PCQ is a 45-45-90 triangle. So
QC = x
And
BQ + QC = 4
Substitute x√3 in for BQ and x in for QC
x√3 + x = 4
Factor x out of the terms on the left side.
x(√3 + 1) = 4
Divide both sides of the equation by (√3 + 1)
x = 4√3+1
Multiply the numerator and denominator by √3 - 1 and simplify
x = 4√3+1⋅√3−1√3−1
x = 4√3−43−1
x = 4√3−42
x = 2√3−2