PQ=DE Given
PB=AE Given
PB-AB=AE-AB substitution
PA=EB Algebra
QA is perpendicular to PE Given
∠PAQ=90 Definition of perpendicular lines
Triangle PAQ is a right triangle Definition of right triangle
DB is perpendicular to PE Given
∠EBD=90 Definition of perpendicular lines
Triangel EBD is a right triangle Definition of right triangle
ΔPAQ is congruent to ΔEBD Hypotenuse-leg Theorem
∠D=∠Q CPCTC