# We have AB=BC, AC=CD, angle ACD=90^circ. If the radius of the circle is 'r' , find BC in terms of r .

We have $AB=BC$, $AC=CD$, $\mathrm{\angle }ACD={90}^{\circ }$. If the radius of the circle is 'r' , find BC in terms of r .
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agyalapi60
Step 1
Angle ABC is $135°$ because it is half the concave angle $COA=270°$.
Step 2
Thus angle $BAC=22.5°$ and $BC=2r\mathrm{sin}22.5°=2r\cdot \frac{\sqrt{2-\sqrt{2}}}{2}=r\sqrt{2-\sqrt{2}}$