Question

in the adjoining figure, AB=AC and AD=AE. Prove that 1. `/_ADB=/_AEC` 2. `DeltaABD~=DeltaACE` 3. BE=DC

Answer 1

`/_ABE=/_ACE `(Isosceles)
`/_ADE=/_AED`(Isosceles)
1)`180-/_ADE=180-/_AED`
`/_ADB=/_AEC`
2)`In/_ABD and /_ACE`
`/_ABD=/_ACE`
`/_ADB=/_AEC`
`/_BAD=/_CAE`
`/_ABD cong /_ACE`(SAS)
`AB=AC`
`/_BAD=/_CAE`
`AD=AE`
`/_ABE and /_ACD`
`AB=AC`
`AE=AD`
`/_BAE=/_CAD`
`/_ABE cong /_ACD`(SAS)
BE=CD
`/_BAD+/_DAE=/_CAE+/_DAE`
`/_BAD=/_CAE`.