+0 n triangle ABC, angle ACB = 90 degrees. Let H be the foot of the altitude from C to side AB.
If BC = 1 and AH = 2, find CH.
+14943 \(h^2=1^2-(2r-2)^2=r^2-(2-r)^2\\ 1-(4r^2-8r+4)=r^2-(4-4r+r^2)\\ 1-4r^2+8r-4=r^2-4+4r-r^2\\ 4r^2-4r-1=0\\ r^2-r-0.25=0\\ \)
\(r=0.5\pm\sqrt{0.25+0.25}\\ r\in\{1.207,-0.207\}\\ \color{blue}r=0.5+\sqrt{0.5}=1.207\\ h^2=1^2-(2r-2)^2\\ h^2=1-(1+\sqrt{2}-2)^2\\ h^2=1-(\sqrt{2}-1)^2\\ h^2=1-(2-2\sqrt{2}+1)\\ h=\sqrt{2\sqrt{2}-2}\\ \color{blue}h=\overline{CH}=0.91\)
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