In triangle ABC, AC = BC, angle DCB is 40 degrees, and CD is parallel to AB. What is the number of degrees in angle ECD?

∠DCB  and  ∠CBA  are alternate interior angles, so they have the same measure.

 

∠CBA  and  ∠BAC  are base angles of isosceles triangle ABC, so they have the same measure.

 

So..

 

m∠DCB   =   m∠CBA   =   m∠BAC   =   40°

 

And since there are  180°  in triangle  ABC,

 

m∠ACB   =   180° - 40° - 40°   =   100°

 

And...

 

m∠ACB  +  m∠DCB  +  m∠ECD   =   180°

 

100°  +  40°  +  m∠ECD   =   180°

 

m∠ECD   =   180°  -  40°  -  100°   =   40°