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Triangles: Isosceles

If $$AD=DB$$ and $$DC=BC$$ as shown above, what is the value of $$xº$$?
Correct. [[snippet]] Given that $$AD = DB$$, $$\angle DAB = \angle DBA = 40º$$. Since the sum of all angles in a triangle is $$180º$$, > $$ADB = 180º - 2(40º) = 180º - 80º = 100º$$. Based on this, angle $$BDC$$, the supplementary angle of $$ADB$$, is $$180º - 100º = 80º$$. Hence > $$xº = 180º - 2(80º) = 180º - 160º = 20º$$.
Incorrect. [[snippet]]
Incorrect. [[snippet]]
Incorrect. [[snippet]]
Incorrect. [[snippet]]
$$20º$$
$$40º$$
$$60º$$
$$80º$$
$$100º$$