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$$\frac{CD}{\sin(CBD)}=\frac{BD}{\sin(BCD)}$$ $$\frac{AD}{\sin(ABD)}=\frac{BD}{\sin(BAD)}$$ $$\sin(BAD)=\frac{BD\sin(ABD)}{AD}\\ =\frac{BD\sin(CBD)}{CD}\\ =\sin(BCD)$$ $$\frac{AB}{\sin(BCD)}=\frac{BC}{\sin(BAD)}$$ $$AB=BC$$

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$$\arctan(a)+\arctan(b)+\arctan(c)=\pi$$

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$$a+b+c+d=1\\a+d=\frac{1}{2}\\a+b=\frac{1}{4}\\c=4a$$ $$a=\frac{1}{12}\\b=\frac{1}{6}\\c=\frac{1}{3}\\d=\frac{5}{12}\\$$

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$$\frac{1}{2}\times 1\times 1\times \sin(60)\\=\frac{\sqrt{3}}{4}$$ $$\frac{1}{2}ab \sin(60)\\=\frac{ab\sqrt{3}}{4}$$ $$ab=\frac{1}{2}$$ $$L=a^2+b^2-2ab\cos(60)\\ =a^2+\left(\frac{1}{2a}\right)^2-\frac{2a}{2a}\times \frac{1}{2}\\ =a^2+\frac{1}{4a^2}-\frac{1}{2}$$ $$0=\frac{dL}{da}=2a-\frac{1}{2a^3}\\ 4a^4-1=0\\ a^4=\frac{1}{4}\\ a=\frac{1}{\sqrt{2}}$$

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$$8^2+r^2=(12-r)^2$$ $$64+r^2=144-24r+r^2$$ $$24r=80$$ $$r=\frac{80}{24}=\frac{10}{3}$$

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