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saul85 [17]
3 years ago
9

Please help !! prove abp = dcp

Mathematics
1 answer:
Ksivusya [100]3 years ago
8 0

Answer:

  1. ∠1≅∠2-------------------------(Given)
  2. m∠1=m∠2-------------------(Definition of congruent angles)
  3. ∠ABP, ∠1 and ∠DCP,∠2 form linear pair---------------Linear pair
  4. m∠ABP+m∠1=180° and m∠DCP+m∠2=180°--------definition of linear pair
  5. ∠ABP≅∠DCP--------------------If equals are subtracted from equals, the remainders are equal
  6. AP≅DP----------------------------Given
  7. ΔABP≅ΔDCP-------------------AAS

Step-by-step explanation:

Given, ∠1≅∠2, ∠3≅∠4 and AP≅DP,

  • We know that,

        ∠1≅∠2(given)

     ⇒m∠1=m∠2(definition of congruent angles)

  • ∠ABP+m∠1=180° and ∠DCP+m∠2=180° (Linear Pair)

         180°=180°(Reflexive)

      ⇒m∠ABP+m∠1=m∠DCP+m∠2

          But  m∠1 =m∠2 (definition of congruent angles)

     ⇒m∠ABP+m∠1=m∠DCP+m∠1

        m∠ABP+=m∠DCP[/tex] ( If equals are subtracted from equals, the remainders are equal)[tex]AP=DP(GIven)

Therefore, ΔABP=ΔDCP (by AAS criteria)

condition are:

  1. m∠1=m∠2 (Angle)
  2. m∠ABP=m∠DCP (Angle)
  3. AP=DP (Side)
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