Let us have a triangle ABC and a point D on BC such that BD=DC=DA. If angle ACB=50 then how many degrees are in angle ABC?
because CD=AD, triangle CAD is iscoscolese, so angle CAD is 50 as well, which makes CDA 80, and because D is on CB, angle ADB+ADC=180, so ADB=80, and because AD=BD, ADB is an iscoscolese triangle, so angle ABC=(180-80)/2=$\boxed{50^\circ}$
ΔABD is an isosceles triangle.
Angle ABD = 100 degrees, and angles ABD and BAD are congruent.
Angle ABC = 1/2(180 - 100) = 40 degrees
