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ADC and ABC are triangles such that $$AD=DC$$ and $$CA=AB$$. If $$\angle CAB=20^\circ$$ and $$\angle ADC=100^\circ$$, without using trigonometry, prove that $$AB=BC+CD$$.
Correct Answer: 0
Extend BC to a point E such that $$CE=CD$$, making triangle CED equilateral since the base angles work out to $$60^\circ$$. Joining AE and tracking the resulting angles shows triangle ABE is isosceles with $$AB=BE$$. Since $$BE=BC+CE=BC+CD$$, this gives $$AB=BC+CD$$.
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