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

1 Answer
Jul 2, 2018

Answer:

#m/_CDA=108^o#

Explanation:

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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#