如圖,設(shè)直角三角形邊長AC=x,BC=y,AD=DE=EB=z    則有AB=3z
由余弦定理可知：CD^2=AD^2+AC^2-2AD*AC*cosA=(sina)^2          式1  cosA=x/3z  (鄰邊比斜邊)
                            CE^2=EB^2+BC^2-2EB*BC*cosB=(cosa)^2          式2  cosB=y/3z(鄰邊比斜邊)
上兩等式相加,代入數(shù)據(jù)則有：2z^2+x^2+y^2-2x^2/3-2y^2/3=1
                               再次化簡：（x^2+y^2)/3+2z^2=1       根據(jù)勾股定理又知：x^2+y^2=AB^2=9z^2
                                                             3z^2+2z^2=1 
解得                                                                    z=(根號5)/5
所以AB=3z=3*(根號5)/5