If $$AD=DB$$ and $$DC=BC$$ as shown above, what is the value of $$xº$$?

Correct.
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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.
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Incorrect.
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Incorrect.
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Incorrect.
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$$20º$$

$$40º$$

$$60º$$

$$80º$$

$$100º$$