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