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.
Determine the measure of angle BPC
+605 There is no diagram, so I am assuming $E\in ABCD$.
We have $\angle PAE=60-\angle BAC=60-45=15$
And $\angle APE=180-60-15=180-75=105$.
By Vertical angles, $\angle APE=\angle BPC=\boxed{105^{\circ}}$.
