Respuesta :

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:

  • ∠DBE=∠QAP=90°(right angle)
  • PQ=ED(hypotenuse)
  • PA=EB(side)

So, ∡D=∡Q(as congruent parts of congruent triangles are equal)

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