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
Teddy
Jul 2, 2018

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

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