Question

In an Isoceles `triangle` `ABC`,bisector `CD` of the `angle` `C` is equal to the base `BC`.Then the angle between `CDA` is ?

Answer 1

`m/_CDA=108^o`

Explanation:

Since `ABC` is isosceles with base `BC`, we know that `AB=AC`. Therefore, `m/_ABC=m/_ACB`. Since we are given that `CD=BC`, we know that `m/_CDB=m/_CBD`. Let `x=m/_DCB`. Since `CD` is the bisector of `/_C`, we know that `m/_BCA=2x=m/_ABC=m/_CDB`. Since `CDB` is a triangle, we know that `m/_CDB+m/_CBD+m/_BCD=108^o=x+2x+2x`.
`5x=180^o`
`:.x=36^o`.
`:.m/_CDB=2x=72^o`
Since `/_CDB` and `/_CDA` are supplementary, `m/_CDB+m/_CDA=180^o`
`:.72^o+m/_CDA=180^o`
`:.m/_CDA=108^o`