In an Isoceles triangle ABCABC,bisector CDCD of the angle CC is equal to the base BCBC.Then the angle between CDACDA is ?

1 Answer
Jul 2, 2018

m/_CDA=108^omCDA=108o

Explanation:

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Since ABCABC is isosceles with base BCBC, we know that AB=ACAB=AC. Therefore, m/_ABC=m/_ACBmABC=mACB. Since we are given that CD=BCCD=BC, we know that m/_CDB=m/_CBDmCDB=mCBD. Let x=m/_DCBx=mDCB. Since CDCD is the bisector of /_CC, we know that m/_BCA=2x=m/_ABC=m/_CDBmBCA=2x=mABC=mCDB. Since CDBCDB is a triangle, we know that m/_CDB+m/_CBD+m/_BCD=108^o=x+2x+2xmCDB+mCBD+mBCD=108o=x+2x+2x.
5x=180^o5x=180o
:.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