Question

Please help !!
prove abp = dcp

Answer 1

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

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</strong>] ( If equals are subtracted from equals, the remainders are equal)</p><ul><li><strong>[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)