非截长补短思想

已知，如图\(BD\perp AC\)，\(\angle AEC=45^\circ\)，\(BD=2CE\)，\(DE\parallel BC\)，求证\(CE=AC+AD\)．

解作\(CF\perp AB\)交\(AB\)于点\(F\)． 设\(CE=a\)，\(AD=x\)，所以\[BD=2a,EF=CF=\dfrac{\sqrt 2}{2}a.\]因为\(\triangle CAF\backsim \triangle BAD\)， 所以\[\dfrac{AC}{AB}=\dfrac{AF}{AD}=\dfrac{CF}{BD}=\dfrac{\sqrt 2}{4},\]所以\[AF=\dfrac{\sqrt 2}{4}x,AC=\sqrt{\dfrac 18x^2+\dfrac 12a^2}.\]因为\(DE\parallel BC\)，所以\[\dfrac{AD}{AC}=\dfrac{AE}{AB},\] 所以\[\begin{split} AD & =\dfrac{AC}{AB}\times AE\\ &=\dfrac{\sqrt 2}{4}(EF+AF)\\ &=\dfrac{\sqrt 2}{4}(\dfrac{\sqrt 2}{2}a+\dfrac{\sqrt 2}{4}x)\\ &=x.\end{split}\]解得\[x=\dfrac 27a,AC=\dfrac 57 a.\]所以\[CE=AD+AC.\]