Question

Please help !! prove abp = dcp

Answer 1

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, Δ$ABP=$Δ$DCP$ (by AAS criteria)

condition are:

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