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in the adjoining figure, AB=AC and AD=AE. Prove that1. `/_ADB=/_AEC`2. `DeltaABD~=DeltaACE`3. BE=DC |
Answer» `/_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`. |
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