In triangle $ABC,$ $D$ is a point on $\overline{BC}$ such that $BD = DC = DA.$ If $\angle ABC = 45^\circ,$ then how many degrees are in $\angle ACB$?
I think the answer is 45 degrees, because angle abc is 45, and because BD = BC = BA, I think it is an isosceles. I'm not sure.
By SAS, triangle BDA is similar to triangle CDA
Angle ABD = Angle ACD = Angle ACB = 45
Another way would be to note that ADB and ADC are congruent, which proves that the angle is 45 as well.