Plz prove this triangle congruence.

Answer:
ΔDBE≅ΔQAP (by RHS criteria)
Step-by-step explanation:
Given that, [tex]PQ=DE[/tex], [tex]PB=AE[/tex], [tex]QA[/tex]⊥[tex]PE[/tex]
and [tex]DB[/tex]⊥[tex]PE[/tex]
⇒∠PAQ=90° and ∠EBD=90°(definition of perpendicular lines)
Its given that PB=AE,
subtracting AB on both sides,
we get: PB-AE=AB-AE
⇒PA=EB (equals subtracted from equals, the remainders are equals)
Therefore, ΔDBE≅ΔQAP (by RHS criteria)
conditions for congruence:
So, ∡D=∡Q(as congruent parts of congruent triangles are equal)